Algorithms by Jeff Erickson

Algorithms

Jeff Erickson

oth edition (pre-publication draft) — December 29, 2018
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© Copyright 2019 Jeff Erickson

This work is available under a Creative Commons Attribution 4.0 International License.
For license details, see http://creativecommons.0rg/licenses/by/4.o/.

Download this book at http://jeffe.cs.illinois.edu/teaching/algorithms/
or http://algorithms.wtf

Please report errors at https://github.com/jeffgerickson/algorithms

Portions of our programming are mechanically reproduced,
and we now begin our broadcast day.

For Kim, Kate, and Hannah
with love and admiration

And for Erin
with thanks

for breaking her promise

Incipit prologus in libro alghoarismi de practica arismetrice.

— loannis Hispalensis [John of Seville?],
Liber algorismi de pratica arismetrice (c.i 135)

Shaii I tell you, my friend, how you will come to understand it?

Go and write a book upon it.

- Henry Home, Lord Kames (1696-1782),
in a letter to to Sir Gilbert Elliot

The individual is always mistaken. He designed many things, and drew in other
persons as coadjutors, quarrelled with some or all, blundered much, and
something is done; all are a little advanced, but the individual is always mistaken.

It turns out somewhat new and very unlike what he promised himself.

- Ralph Waldo Emerson, "Experience", Essays, Second Series (1844)

What I have outlined above is the content of a book the realization of whose basic
plan and the incorporation of whose details would perhaps be impossible; what I
have written is a second or third draft of a preliminary version of this book

— Michael Spivak, preface of the first edition of
Differential Geometry, Volume I (1970)

Preface

About This Book

This textbook grew out of a collection of lecture notes that I wrote for various
algorithms classes at the University of Illinois at Urbana-Champaign, which I
have been teaching about once a year since January 1999. Spurred by changes
of our undergraduate theory curriculum, I undertook a major revision of my
notes in 2016; this book consists of a subset of my revised notes on the most
fundamental course material, mostly reflecting the algorithmic content of our
new required junior-level theory course.

Prerequisites

The algorithms classes I teach at Illinois have two significant prerequisites:
a course on discrete mathematics and a course on fundamental data structures.
Consequently, this textbook is probably not suitable for most students as a first

Preface

course in data structures and algorithms. In particular, I assume at least passing
familiarity with the following specific topics:

• Discrete mathematics: High-school algebra, logarithm identities, naive
set theory, Boolean algebra, first-order predicate logic, sets, functions,
equivalences, partial orders, modular arithmetic, recursive definitions, trees
(as abstract objects, not data structures), graphs (vertices and edges, not
function plots).

• Proof techniques: direct, indirect, contradiction, exhaustive case analysis,
and induction (especially “strong” and “structural” induction). Chapter o
uses induction, and whenever Chapter n— 1 uses induction, so does Chapter n.

• Iterative programming concepts: variables, conditionals, loops, records,
indirection (addresses/pointers/references), subroutines, recursion. I do not
assume fluency in any particular programming language, but I do assume
experience with at least one language that supports both indirection and
recursion.

• Fundamental abstract data types: scalars, sequences, vectors, sets, stacks,
queues, maps/dictionaries, ordered maps/dictionaries, priority queues.

• Fundamental data structures: arrays, linked lists (single and double,
linear and circular), binary search trees, at least one form of balanced binary
search tree (such as AVL trees, red-black trees, treaps, skip lists, or splay
trees), hash tables, binary heaps, and most importantly, the difference
between this list and the previous list.

• Fundamental computational problems: elementary arithmetic, sorting,
searching, enumeration, tree traversal (preorder, inorder, postorder, level-
order, and so on).

• Fundamental algorithms: elementary algorism, sequential search, binary
search, sorting (selection, insertion, merge, heap, quick, radix, and so
on), breadth- and depth-first search in (at least binary) trees, and most
importantly, the difference between this list and the previous list.

• Elementary algorithm analysis: Asymptotic notation (o, O, 0 , Q, o>),
translating loops into sums and recursive calls into recurrences, evaluating
simple sums and recurrences.

• Mathematical maturity: facility with abstraction, formal (especially recur¬
sive) definitions, and (especially inductive) proofs; writing and following
mathematical arguments; recognizing and avoiding syntactic, semantic,
and/or logical nonsense.

The book briefly covers some of this prerequisite material when it arises in
context, but more as a reminder than a good introduction. For a more thorough
overview, I strongly recommend the following freely available references:

Additional References

• Margaret M. Fleck. Building Blocks for Theoretical Computer Science, un¬
published textbook, most recently revised January 2013. Available from
http ://mfleck. cs. illinois. edu/building- blocks/.

• Eric Lehman, F. Thomson Leighton, and Albert R. Meyer. Mathematics for
Computer Science, unpublished lecture notes, most recent (public) revision
June 2018. Available from https://courses.csail.mit.edU/6.042/spring18/.
(I strongly recommend searching for the most recent revision.)

• Pat Morin. Open Data Structures, most recently revised January 2016 (edition
o.iG/ 3 ). A free open-content textbook, which Pat maintains and regularly
updates. Available from http://opendatastructures.org/.

Additional References

Please do not restrict yourself to this or any other single reference. Authors and
readers bring their own perspectives to any intellectual material; no instructor
“clicks” with every student, or even with every very strong student. Finding the
author that most effectively gets their intuition into your head takes some effort,
but that effort pays off handsomely in the long run.

The following references have been particularly valuable sources of intuition,
examples, exercises, and inspiration; this is not meant to be a complete list.

• Alfred V. Aho, John E. Hopcroft, and Jeffrey D. Ullman. The Design and
Analysis of Computer Algorithms. Addison-Wesley, 1974. (I used this textbook
as an undergraduate at Rice and again as a masters student at UC Irvine.)

• Thomas Cormen, Charles Leiserson, Ron Rivest, and Cliff Stein. Introduction
to Algorithms, third edition. MIT Press/McGraw-Hill, 2009. (I used the first
edition as a teaching assistant at Berkeley.)

• Sanjoy Dasgupta, Christos H. Papadimitriou, and Umesh V. Vazirani. Algo¬
rithms. McGraw-Hill, 2006. (Probably the closest in content to this book,
but considerably less verbose.)

• Jeff Edmonds. How to Think about Algorithms. Cambridge University Press,
2008.

• Michael R. Garey and David S. Johnson. Computers and Intractability:
A Guide to the Theory of NP-Completeness. W. H. Freeman, 1979.

• Michael T. Goodrich and Roberto Tamassia. Algorithm Design: Foundations,
Analysis, and Internet Examples. John Wiley & Sons, 2002.

• Jon Kleinberg and Eva Tardos. Algorithm Design. Addison-Wesley, 2005.
Borrow it from the library if you can.

• Donald Knuth. The Art of Computer Programming, volumes 1-4A. Addison-
Wesley, 1997 and 2011. (My parents gave me the first three volumes for
Christmas when I was 14. Alas, I didn’t actually read them until much later.)

Preface

• Udi Manber. Introduction to Algorithms: A Creative Approach. Addison-
Wesley, 1989. (I used this textbook as a teaching assistant at Berkeley.)

• Ian Parberry. Problems on Algorithms. Prentice-Hall, 1995 (out of print).
Downloadable from https://larc.unt.edu/ian/books/free/license.html after
you agree to make a small charitable donation. Please honor your agreement.

• Robert Sedgewick and Kevin Wayne. Algorithms. Addison-Wesley, 2011.

• Robert Endre Tarjan. Data Structures and Network Algorithms. SIAM, 1983.

• Class notes from my own algorithms classes at Berkeley, especially those
taught by Dick Karp and Raimund Seidel.

• Lecture notes, slides, homeworks, exams, video lectures, research papers,
blog posts, and full-fledged MOOCs made freely available on the web by
innumerable colleagues around the world.

About the Exercises

Each chapter ends with several exercises, most of which I have used at least
once in a homework assignment, discussion/lab section, or exam. The exercises
are not ordered by increasing difficulty, but (generally) clustered by common
techniques or themes. Some problems are annotated with symbols as follows:

• *Red hearts indicate particularly challenging problems; many of these have
appeared on qualifying exams for PhD students at Illinois. A small number
of really hard problems are marked with ^larger hearts, and a few open
problems are marked with Venormous hearts.

• *Blue diamonds indicate problems that require familiarity with material
from later chapters, but thematically belong where they are. Problems that
require familiarity with earlier material are not marked, however; the book,
like life, is cumulative.

• ’’’Green clubs indicate problems that require familiarity with material out¬
side the scope of this book, such as finite-state machines, linear algebra,
probability, or planar graphs. These are rare.

• *Black spades indicate problems that require a significant amount of grunt
work and/or coding. These are rare.

• * Orange stars indicate that you are eating Lucky Charms that were manu¬
factured before 1998. Ew.

These exercises are designed as opportunities to practice, not as targets for their
own sake; the goal of these problems not to solve these particular problems,
but to practice exercising a particular skill, or solving a type of problem. Partly
for this reason, I don’t provide solutions to the exercises; the solutions are not
the point. In particular, there is no “instructor’s manual”; if you can’t solve a

IV

Steal This Book!

problem yourself, you probably shouldn’t assign it to your students. That said,
you can probably find solutions to whatever homework problems I’ve assigned
this semester on the web page of whatever course I’m teaching. And nothing is
stopping you from writing an instructor’s manual!

Steal This Book!

This book is published under a Creative Commons Licence that allows you to use,
redistribute, adapt, and remix its contents without my permission, as long as you
point back to the original source. A complete electronic version of this book is
freely available at my web site http://jeffe.cs.illinois.edu/teaching/algorithms/
(or the mnemonic shortcut http://algorithms.wtf), at the bug-report site https://
github.com/jeffgerickson/algorithms, and on the Internet Archive (ask Google).

The book web site also contains several hundred pages of additional lecture
notes on related and more advanced material, as well as a near-complete
archive of past homeworks, exams, discussion/lab problems, and other teaching
resources. Whenever I teach an algorithms class, I revise, update, and sometimes
cull my teaching materials, so you may find more recent revisions on the web
page of whatever course I am currently teaching.

Whether you are a student or an instructor, you are more than welcome to use
any subset of this textbook or my other lecture notes in your own classes, without
asking my permission—that’s why I put them on the web! However, please also
cite this book, either by name or with a link back to http://algorithms.wtf; this
is especially important if you are a student, and you use my course materials to
help with your homework. (Please also check with your instructor.)

However, if you are an instructor, I strongly encourage you to supplement
these with additional material that you write yourself. Writing the material
yourself will strengthen your mastery and in-class presentation of the material,
which will in turn improve your students’ mastery of the material. It will also
get you past the frustration of dealing with the parts of this book that you don’t
like. All textbooks are crap imperfect, and this one is no exception.

Finally, please make whatever you write freely, easily, and globally avail¬
able on the open web —not hidden behind the gates of a learning management
system—so that students and instructors elsewhere can benefit from your unique
insights. In particular, if you develop useful resources that directly complement
this textbook, such as slides, videos, or solution manuals, please let me know so
that I can add links to your resources from the book web site.

v

Preface

Acknowledgments

This textbook draws heavily on the contributions of countless algorithms students,
teachers, and researchers. In particular, I am immensely grateful to more than
three thousand Illinois students who have used my lecture notes as a primary
reference, offered useful (if sometimes painful) criticism, and suffered through
some truly awful early drafts. Thanks also to many colleagues and students
around the world who have used these notes in their own classes and have sent
helpful feedback and bug reports.

I am particularly grateful for the feedback and contributions (especially
exercises) from my amazing teaching assistants:

Aditya Ramani, Akash Gautam, Alex Steiger, Alina Ene, Amir Nayyeri,

Asha Seetharam, Ashish Vulimiri, Ben Moseley, Brad Sturt, Brian Ensink,

Chao Xu, Charlie Carlson, Chris Neihengen, Connor Clark, Dan Bullok,

Dan Cranston, Daniel Khashabi, David Morrison, Ekta Manaktala, Erin
Wolf Chambers, Gail Steitz, Gio Kao, Grant Czajkowski, Hsien-Chih Chang,

Igor Gammer, Jacob Laurel, John Lee, Johnathon Fischer, Junqing Deng,

Kent Quanrud, Kevin Milans, Kevin Small, Konstantinos Koiliaris, Kyle Fox,

Kyle Jao, Lan Chen, Mark Idleman, Michael Bond, Mitch Harris, Naveen
Arivazhagen, Nick Bachmair, Nick Hurlburt, Nirman Kumar, Nitish Korula,
Patrick Lin, Phillip Shih, Rachit Agarwal, Reza Zamani-Nasab, Rishi Talreja,

Rob McCann, Sahand Mozaffari, Shalan Naqvi, Shripad Thite, Spencer
Gordon, Srihita Vatsavaya, Subhro Roy, Tana Wattanawaroon, Umang
Mathur, Vipul Goyal, Yasu Furakawa, and Yipu Wang.

I’ve also been helped tremendously by many discussions with faculty col¬
leagues at Illinois: Alexandra Kolia, Cinda Heeren, Edgar Ramos, Herbert
Edelsbrunner, Jason Zych, Kim Whittlesey, Lenny Pitt, Madhu Parasarathy,
Mahesh Viswanathan, Margaret Fleck, Shang-Hua Teng, Steve LaValle, and
especially Chandra Chekuri, Ed Reingold, and Sariel Har-Peled.

Of course this book owes a great debt to the people who taught me this
algorithms stuff in the first place: Bob Bixby and Michael Pearlman at Rice;
David Eppstein, Dan Hirschberg, and George Lueker at Irvine; and Abhiram
Ranade, Dick Karp, Manuel Blum, Mike Luby, and Raimund Seidel at Berkeley.

I stole the first iteration of the overall course structure, and the idea to write
up my own lecture notes in the first place, from Herbert Edelsbrunner; the idea
of turning a subset of my notes into a book from Steve LaValle; and several
components of the book design from Robert Ghrist.

Caveat Lector!

VI

Of course, none of those people should be blamed for any flaws in the resulting
book. Despite many rounds of revision and editing, this book contains many

Caveat Lector!

mistakes, bugs, gaffes, omissions, snafus, kludges, typos, mathos, grammaros,
thinkos, brain farts, poor design decisions, historical inaccuracies, anachronisms,
inconsistencies, exaggerations, dithering, blather, distortions, oversimplification,
nonsense, garbage, cruft, junk, and outright lies, all of which are entirely
Steve Skiena’s fault.

I maintain an issue tracker at https://github.com/jeffgerickson/algorithms,
where readers like you can submit bug reports, feature requests, and general
feedback on the book. Please let me know if you find an error of any kind,
whether mathematical, grammatical, historical, typographical, cultural, or
otherwise, whether in the main text, in the exercises, or in my other course
materials. (Steve is unlikely to care.) Of course, all other feedback is also
welcome!

Enjoy!

— Jeff

It is traditional for the author to magnanimously accept the blame for whatever
deficiencies remain. I don't. Any errors, deficiencies, or problems in this book are
somebody else's fault, but i would appreciate knowing about them so as to
determine who is to blame.

— Steven S. Skiena, The Algorithm Design Manual (1997)

No doubt this statement will be followed by an annotated list of all textbooks,
and why each one is crap.

- Adam Contini, MetaFilter, January 4, 2010

VII

Table of Contents

Preface i

About This Book. i

Prerequisites. i

Additional References. iii

About the Exercises. iv

Steal This Book!. v

Acknowledgments. vi

Caveat Lector! . vi

Table of Contents ix

o Introduction 1

o.i What is an algorithm?. 1

0.2 Multiplication. 3

Table of Contents

Lattice Multiplication • Duplation and Mediation • Compass and Straight¬
edge

0.3 Congressional Apportionment. 8

0.4 A Bad Example. to

0.5 Describing Algorithms. 11

Specifying the Problem • Describing the Algorithm

0.6 Analyzing Algorithms. T4

Correctness • Running Time

Exercises. T7

1 Recursion 21

r.i Reductions. 2t

r.2 Simplify and Delegate. 22

1.3 Tower of Hanoi. 24

r.4 Mergesort . 26

Correctness • Analysis

T.5 Quicksort. 29

Correctness • Analysis

r.6 The Pattern. 3t

1.7 Recursion Trees. 31

^Ignoring Floors and Ceilings Is Okay, Honest

r.8 r Linear-Time Selection. 35

Analysis • Sanity Checking

T.9 Fast Multiplication. 39

r.ro Exponentiation. 42

Exercises. 43

2 Backtracking 69

2.r N Queens. 69

2.2 Game Trees . 72

2.3 Subset Sum . 74

Correctness • Analysis • Variants

2.4 The General Pattern. 77

2.5 String Segmentation (Interpunctio Verborum ). 78

Index Formulation • v Analysis • Variants

2.6 Longest Increasing Subsequence. 84

2.7 Longest Increasing Subsequence, Take 2. 87

2.8 Optimal Binary Search Trees. 89

^Analysis

Exercises. 9t

3 Dynamic Programming 95

x

Table of Contents

3.1 Matravrtta. 95

Backtracking Can Be Slow • Memo(r)ization: Remember Everything • Dy¬
namic Programming: Fill Deliberately • Don’t Remember Everything After
All

3.2 r Aside: Even Faster Fibonacci Numbers. 101

Whoa! Not so fast!

3.3 Interpunctio Verborum Redux .103

3.4 The Pattern: Smart Recursion.103

3.5 Warning: Greed is Stupid.105

3.6 Longest Increasing Subsequence.107

First Recurrence: Is This Next? • Second Recurrence: What’s Next?

3.7 Edit Distance.109

Recursive Structure • Recurrence • Dynamic Programming

3.8 Subset Sum . 114

3.9 Optimal Binary Search Trees. 115

3.10 Dynamic Programming on Trees. 118

Exercises. 121

4 Greedy Algorithms 157

4.1 Storing Files on Tape. 157

4.2 Scheduling Classes.159

4.3 General Structure. 162

4.4 Huffman Codes.163

4.5 Stable Matching.168

Some Bad Ideas • The Boston Pool and Gale-Shapley Algorithms • Running
Time • Correctness • Optimality!

Exercises. 174

5 Basic Graph Algorithms 185

5.1 Introduction and History.185

5.2 Basic Definitions.188

5.3 Representations and Examples.190

5.4 Data Structures.193

Adjacency Lists • Adjacency Matrices • Comparison

5.5 Whatever-First Search.197

Analysis

5.6 Important Variants.199

Stack: Depth-First • Queue: Breadth-First • Priority Queue: Best-
First • Disconnected Graphs • Directed Graphs

5.7 Graph Reductions.203

Flood Fill

Exercises.205

XI

Table of Contents

6 Depth-First Search 223

6.1 Preorder and Postorder.225

Classifying Vertices and Edges

6.2 Detecting Cycles.229

6.3 Topological Sort.230

Implicit Topological Sort

6.4 Memoization and Dynamic Programming.232

Dynamic Programming in Dags

6.5 Strong Connectivity.235

6.6 Strong Components in Linear Time.236

Koraraju and Sharir’s Algorithm • r Tarjan’s Algorithm
Exercises.242

7 Minimum Spanning Trees 255

7.1 Distinct Edge Weights.255

7.2 The Only Minimum Spanning Tree Algorithm .257

7.3 Boruvka’s Algorithm.259

This is the MST Algorithm You Want

7.4 Jarnik’s (“Prim’s”) Algorithm. 261

^Improving Jarnik’s Algorithm

7.5 Kruskal’s Algorithm.263

Exercises.265

8 Shortest Paths 271

8.1 Shortest Path Trees. 272

8.2 ^Negative Edges. 272

8.3 The Only SSSP Algorithm.274

8.4 Unweighted Graphs: Breadth-First Search.276

8.5 Directed Acyclic Graphs: Depth-First Search.280

8.6 Best-First: Dijkstra’s Algorithm.283

No Negative Edges • ^Negative Edges

8.7 Relax ALL the Edges: Bellman-Ford.287

Moore’s Improvement • Dynamic Programming Formulation
Exercises.295

9 All-Pairs Shortest Paths 307

9.1 Introduction.307

9.2 Lots of Single Sources.308

9.3 Reweighting.309

9.4 Johnson’s Algorithm.310

9.5 Dynamic Programming . 311

9.6 Divide and Conquer.313

XII

Table of Contents

9.7 Funny Matrix Multiplication.314

9.8 (Kleene-Roy-)Floyd-Warshall(-Ingerman) .316

Exercises.318

10 Maximum Flows & Minimum Cuts 325

10.1 Flows.326

10.2 Cuts.327

10.3 The Maxflow-Mincut Theorem.329

10.4 Ford and Fulkerson’s augmenting-path algorithm.332

^Irrational Capacities

10.5 Combining and Decomposing Flows .334

10.6 Edmonds and Karp’s Algorithms.338

Fattest Augmenting Paths • Shortest Augmenting Paths

10.7 Further Progress .341

Exercises.342

11 Applications of Flows and Cuts 351

11.1 Edge-Disjoint Paths. 351

11.2 Vertex Capacities and Vertex-Disjoint Paths.352

11.3 Bipartite Matching.353

11.4 Tuple Selection.354

Exam Scheduling

11.5 Disjoint-Path Covers.357

Minimal Teaching Assignment

11.6 Baseball Elimination.360

11.7 Project Selection .363

Exercises.365

12 NP-Hardness 377

12.1 A Game You Can’t Win.377

12.2 P versus NP.379

12.3 NP-hard, NP-easy, and NP-complete.380

12.4 ^Formal Definitions (HC SVNT DRACONES ).382

12.5 Reductions and Sat.383

12.6 3 Sat (from Sat) .386

12.7 Maximum Independent Set (from 3 Sat) .388

12.8 The General Pattern.390

12.9 Clique and Vertex Cover (from Independent Set) .391

12.10 Graph Coloring (from 3 Sat) .392

12.11 Hamiltonian Cycle.395

From Vertex Cover • From 3Sat • Variants and Extensions

12.12 Subset Sum (from Vertex Cover).400

xiii

Table of Contents

Caveat Reductor!

12.13 Other Useful NP-hard Problems.402

12.14 Choosing the Right Problem.405

12.15 A Frivolous Real-World Example.406

12.16 y On Beyond Zebra.409

Polynomial Space • Exponential Time • Excelsior!

Exercises.412

Image Credits 427

Colophon 429

XIV

Hinc incipit algorismus. Haec algorismus ars praesens dicitur in qua
talibus indorum fruimur bis quinque figuris 0. g. 8. 7. 6. 5. 4. 3. 2. t.

— Friar Alexander de Villa Dei, Carmen de Algorismo (c. 1220)

You are right to demand that an artist engage his work consciously,

but you confuse two different things:

solving the problem and correctly posing the question.

— Anton Chekhov, in a letter to A. S. Suvorin (October 27,1888)

The more we reduce ourselves to machines in the lower things,
the more force we shall set free to use in the higher.

— Anna C. Brackett, The Technique of Rest (1892)

And here I am at 2:30 a.m. writing about technique, in spite of a strong conviction
that the moment a man begins to talk about technique that's proof that he is fresh
out of ideas.

— Raymond Chandler, letter to Erie Stanley Gardner (May 5,1939)

Good men don't need rules.

Today is not the day to find out why I have so many,

— The Doctor [Matt Smith], "A Good Man Goes to War", Doctor Who (2011)

Introduction

0.1 What is an algorithm?

An algorithm is an explicit, precise, unambiguous, mechanically-executable
sequence of elementary instructions, usually intended to accomplish a specific
purpose. For example, here is an algorithm for singing that annoying song “99
Bottles of Beer on the Wall”, for arbitrary values of 99:

BOTTLESOFBEER(n):

For i <— n down to 1

Sing “i bottles of beer on the wall, i bottles of beer"

Sing “Take one down, pass it around, i — 1 bottles of beer on the wall"

Sing “No bottles of beer on the wall, no bottles of beer"

Sing “Go fo the store, buy some more, n bottles of beer on the wall."

The word “algorithm” does not derive, as algorithmophobic classicists might
guess, from the Greek roots arithmos (uqiO/ioc), meaning “number”, and algos

1

o. Introduction

(d/Vyoc:), meaning “pain”. Rather, it is a corruption of the name of the 9 th century
Persian mathematician Muhammad ibn Musa al-Khwarizml. 1 Al-Khwarizml is
perhaps best known as the writer of the treatise Al-Kitab al-mukhtasar fihisab
al-gabr wa’l-muqabala, 2 from which the modern word algebra derives. In a
different treatise, al-Khwarizml described the modern decimal system for writing
and manipulating numbers—in particular, the use of a small circle or sifr to
represent a missing quantity—which had been developed in India several
centuries earlier. The methods described In Al-Kitab, using either written figures
or counting stones, became known in English as algorism or augrym, and its
figures became known in English as ciphers.

Although both place-value notation and al-KhwarizmFs works were already
known by some European scholars, the “Hindu-Arabic” numeric system was
popularized in Europe by the medieval Italian mathematician and tradesman
Leonardo of Pisa, better known as Fibonacci. Thanks in part to his 1202 book
Liber Abaci, 3 written figures began to replace the counting table (then known
as an abacus ) and finger arithmetic 4 as the preferred platform for calculation 5
in Europe in the 13th century— not because written decimal figures were any
easier to learn or use, but because they provided an audit trail. Ciphers became
common in Western Europe only with the advent of movable type, and truly
ubiquitous only after cheap paper became plentiful in the early 19th century.

Eventually the word algorism evolved into the modern algorithm, via folk
etymology from the Greek arithmos (and perhaps the previously mentioned
algos). 6 Thus, until very recently, the word algorithm referred exclusively

“‘Mohammad, father of Adbdulla, son of Moses, the Kwarizmian”. Kwarizm is an ancient
city, now called Khiva, in the Khorezm Province of Uzbekistan.

“The Compendious Book on Calculation by Completion and Balancing”

3 While it is tempting to translate the title Liber Abaci as “The Book of the Abacus”, a more
accurate translation is “The Book of Calculation”. Both before and after Fibonacci, the Italian
word abaco was used to describe anything related to numerical calculation—devices, methods,
schools, books, and so on—much in the same way that “computer science” is used today in
English, or as the Chinese phrase for “operations research” translates literally as “the study of
using counting rods”.

Reckoning with digits'. ■*»

5 The word calculate derives from the Latin word calculus, meaning “small rock”, referring to
the stones on a counting table, or as Chaucer called them, augrym stones. In 440BC, Herodotus
wrote in his Histories (II.36) that “The Greeks write and calculate (/toY^soBai tphcpoig, literally
“reckon with pebbles”) from left to right; the Egyptians do the opposite. Yet they say that their
way of writing is toward the right, and the Greek way toward the left.” (Herodotus is strangely
silent on which end of the egg the Egyptians ate first.)

6 Some medieval sources claim that the Greek prefix “algo-” means “art” or “introduction”.
Others claim that algorithms were invented by a Greek philosopher, or a king of India, or perhaps
a king of Spain, named “Algus” or “Algor” or “Argus”. A few, possibly including Dante Alighieri,
even identified the inventor with the mythological Greek shipbuilder and eponymous argonaut.
It’s unclear whether any of these risible claims were intended to be historically accurate, or
merely mnemonic.

2

0.2. Multiplication

to mechanical techniques for place-value arithmetic using “Arabic” numerals.
People trained in the fast and reliable execution of these procedures were called
algorists or computators, or more simply, computers.

0.2 Multiplication

Although they have been a topic of formal academic study for only a few decades,
algorithms have been with us since the dawn of civilization. Descriptions of
step-by-step arithmetic computation are among the earliest examples of written
human language, long predating the expositions by Fibonacci and al-Khwarizmi,
or even the place-value notation they popularized.

Lattice Multiplication

The most familiar method for multiplying large numbers, at least for American
students, is the lattice algorithm. This algorithm was popularized by Fibonacci
in Liber Abaci, who learned it from Arabic sources including al-Khwarizrm, who
in turn learned it from Indian sources including Brahmagupta’s 7th-century
treatise Brahmasphutasiddhanta, who may have learned it from Chinese sources.
The oldest surviving descriptions of the algorithm appear in The Mathematical
Classic of Sunzi, written in China between the 3rd and 5th centuries, and in
Eutocius of Ascalon’s commentaries on Archimedes’ Measurement of the Circle,
written around 500CE, but there is evidence that the algorithm was known much
earlier. Eutocius credits the method to a lost treatise of Apollonius of Perga,
who lived around 300BCE, entitled Okytokion (’Qkutokiov). 7 The Sumerians
recorded multiplication tables on clay tablets as early as 26oobce, suggesting
that they may have used the lattice algorithm. 8

The lattice algorithm assumes that the input numbers are represented as
explicit strings of digits; I’ll assume here that we’re working in base ten, but the
algorithm generalizes immediately to any other base. To simplify notation, 9 the

7 Literally “medicine that promotes quick and easy childbirth”! Pappus of Alexandria repro¬
duced several excerpts of Okytokion about 200 years before Eutocius, but his description of the
lattice multiplication algorithm (if he gave one) is also lost.

s There is ample evidence that ancient Sumerians calculated accurately with extremely
large numbers using their base -60 place-value numerical system, but I am not aware of any
surviving record of the actual methods they used. In addition to standard multiplication
and reciprocal tables, tables listing the squares of integers from 1 to 59 have been found,
leading some math historians to conjecture that Babylonians multiplied using an identity like
xy = ((x + y ) 2 — x 2 — y 2 )/2. But this trick only works when x + y < 60; history is silent on how
the Babylonians might have computed x 2 when x > 60.

9 but at the risk of inflaming the historical enmity between Greece and Egypt, or Lilliput and
Blefuscu, or Macs and PCs, or people who think zero is a natural number and people who are
wrong

3

o. Introduction

input consists of a pair of arrays X[0.. m — 1] and T[0.. n — 1], representing the
numbers

m— 1 n—1

x = J]x[i]-10 i and y = J] Y [j]-10 j ,

i=0 j=0

and similarly, the output consists of a single array Z\ 0 .. m + n— 1 ], representing
the product

m+n— 1

z = x ■ y = J] ZM-IO*.
k =o

The algorithm uses addition and single-digit multiplication as primitive opera¬
tions. Addition can be performed using a simple for-loop. In practice, single-digit
multiplication is performed using a lookup table, either carved into clay tablets,
painted on strips of wood or bamboo, written on paper, stored in read-only
memory, or memorized by the computator. The entire lattice algorithm can be
summarized by the formula

m —1 n— 1

t =0 ;=0

Different variants of the lattice algorithm evaluate the partial products X[i] ■
Y[j] • 10 i+; in different orders and use different strategies for computing their
sum. For example, in Liber Abaco, Fibonacci describes a variant that considers
the mn partial products in increasing order of significance, as shown in modern
pseudocode below.

FibonacciMultiply(X|" 0 .. m —

ll,H0..n—11):

hold <— 0

for k <— 0 to n + m — 1

for all i and j such that

i+j = k

hold <— hold + X[i]

Y[j ]

Z[k] <— hold mod 10

hold <— [hold/10\

return Z[ 0 ..m + n— 1 ]

Fibonacci’s algorithm is often executed by storing all the partial products in a
two-dimensional table (often called a “tableau” or “grate” or “lattice”) and then
summing along the diagonals with appropriate carries, as shown on the right in
Figure o.i. American elementary-school students are taught to multiply one
factor (the “multiplicand”) by each digit in the other factor (the “multiplier”),
writing down all the multiplicand-by-digit products before adding them up, as
shown on the left in Figure o.i. This was also the method described by Eutocius,
although he fittingly considered the multiplier digits from left to right, as shown

4

0.2. Multiplication

in Figure 0.2. Both of these variants (and several others) are described and
illustrated side by side in the anonymous 1458 textbook L’Arte dell’Abbaco, also
known as the Treviso Arithmetic, the first printed mathematics book in the West.

9 3 4
% I 4

1

5

i ’3 ^

9 3 4
% $ o z

Z 7 6 j Z

195

O

2* *

I X /, 0 /

/*'/ 9

T7I

r®/

/9

T7

/*

0 /

/

0 /

/4

* /I • /
/*■ f /«[

Figure 0 . 1 . Computing 934 x 314 = 293276 using "long" multiplication (with error-checking by casting
out nines) and “lattice' 1 multiplication, from LArte dell’Abbaco (1458). (See Image Credits at the end of
the book.)

,, ‘"*°K

STtl U pop T]

Q « C
MMM^qtls

OfLOV co o g | d"

11721
X 11721
1000000 )
100000 1
72126 J
100000 1
172121/
77000 1

4900 \

1481 lJ

summa 187387714-

Figure 0 . 2 . Eutocius's 6th-century calculation of 1178^ x 1178^ = 1373877^, in his commentary on
Archimedes' Measurement of the Circle, transcribed (left) and translated into modern notation (right) by
Johan Heiberg (18gi). (See Image Credits at the end of the book.)

All of these variants of the lattice algorithm—and other similar variants
described by Sunzi, al-Khwarizmi, Fibonacci, LArte dellAbbaco, and many other
sources—compute the product of any m-digit number and any n-digit number
in O(mn) time; the running time of every variant is dominated by the number
of single-digit multiplications.

Duplation and Mediation

The lattice algorithm is not the oldest multiplication algorithm for which we
have direct recorded evidence. An even older and arguably simpler algorithm,
which does not rely on place-value notation, is sometimes called Russian peasant
multiplication, Ethiopian peasant multiplication, or just peasant multiplication.

5

o. Introduction

A variant of this algorithm was copied into the Rhind papyrus by the Egyptian
scribe Ahmes around 1650 BC, from a document he claimed was (then) about
350 years old. 10 This algorithm was still taught in elementary schools in Eastern
Europe in the late 20th century; it was also commonly used by early digital
computers that did not implement integer multiplication directly in hardware.

The peasant multiplication algorithm reduces the difficult task of multiplying
arbitrary numbers to a sequence of four simpler operations: (1) determining
parity (even or odd), (2) addition, ( 3 ) duplation (doubling a number), and ( 4 )
mediation (halving a number, rounding down).

PeasantMultiply(x, y):

prod <— 0
while x > 0
if x is odd

prod <— prod + y
x <- Lx/2J

y^y+y

return prod

X

y

prod

0

123

+ 456 =

456

61

+ 912 =

1368

30

4824

15

+ 3648 =

5016

7

+ 7296 =

12312

3

+14592 =

26904

1

+ 29184 =

56088

Figure 0 . 3 . Multiplication by duplation and mediation

The correctness of this algorithm follows by induction from the following
recursive identity, which holds for all non-negative integers x and y:

0

if x = 0

x-y =

• L*/2j-(y + y)

\x/2 \-{y +y) + y

if x is even
if x is odd

Arguably, this recurrence is the peasant multiplication algorithm!

As stated, PeasantMultiply performs O(logx) parity, addition, and media¬
tion operations, but we can improve this bound to 0(logmin{x, y }) by swapping
the two arguments when x > y. Assuming the numbers are represented us¬
ing any reasonable place-value notation (like binary, decimal, Babylonian
hexagesimal, Egyptian duodecimal, Roman numeral, Chinese counting rods,
bead positions on an abacus, and so on), each operation requires at most
0(log(xy)) = 0(logmax{x,y}) single-digit operations, so the overall running
time of the algorithm is 0(logmin{x,y} • logmax{x,y}) = 0(logx • logy).

In other words, this algorithm requires 0(mn) time to multiply an m-digit
number by an n-digit number; up to constant factors, this is the same running

“The version of this algorithm actually used in ancient Egypt does not use mediation or
parity, but it does use comparisons. To avoid halving, the algorithm pre-computes two tables
by repeated doubling: one containing all the powers of 2 not exceeding x, the other containing
the same powers of 2 multiplied by y. The powers of 2 that sum to x are then found by greedy
subtraction, and the corresponding entries in the other table are added together to form the
product.

0.2. Multiplication

time as the lattice algorithm. This algorithm requires (a constant factor!) more
paperwork to execute by hand than the lattice algorithm, but the necessary
primitive operations are arguably easier for humans to perform. In fact, the two
algorithms are equivalent when numbers are represented in binary.

Compass and Straightedge

Classical Greek geometers identified numbers (or more accurately, magnitudes')
with line segments of the appropriate length, which they manipulated using two
simple mechanical tools—the compass and the straightedge—versions of which
had already been in common use by surveyors, architects, and other artisans for
centuries. Using only these two tools, these scholars reduced several complex
geometric constructions to the following primitive operations, starting with one
or more identified reference points.

• Draw the unique line passing through two distinct identified points.

• Draw the unique circle centered at one identified point and passing through

another.

• Identify the intersection point (if any) of two lines.

• Identify the intersection points (if any) of a line and a circle.

• Identify the intersection points (if any) of two circles.

In practice, Greek geometry students almost certainly drew their constructions
on an abax (a( 3 a'|), a table covered in dust or sand. 11 Centuries earlier, Egyptian
surveyors carried out many of the same constructions using ropes to determine
straight lines and circles on the ground. 12 However, Euclid and other Greek
geometers presented compass and straightedge constructions as precise mathe¬
matical abstractions —points are ideal points; lines are ideal lines; and circles
are ideal circles.

Figure 0.4 shows an algorithm, described in Euclid’s Elements about 2500
years ago, for multiplying or dividing two magnitudes. The input consists of
four distinct points A, B, C,D, and the goal is to construct a point Z such that
| AZ\ = \AC | \AD j /\AB\. In particular, if we define \AB\ to be our unit of length,
then the algorithm computes the product of |AC| and |AD|.

Notice that Euclid first defines a new primitive operation RightAngle by
(as modern programmers would phrase it) writing a subroutine. The correctness
of the algorithm follows from the observation that triangles ACE and AZF are

u The written numerals 1 through 9 were known in Europe at least two centuries before
Fibonacci’s Liber Abaci as “gobar numerals”, from the Arabic word ghubar meaning dust, ultimately
referring to the Indian practice of performing arithmetic on tables covered with sand. The Greek
word a(3a| is the origin of the Latin abacus, which also originally referred to a sand table.

“Remember what “geometry” means? Democritus would later refer to these Egyptian
surveyors, somewhat derisively, as arpedonaptai (agTrsbovdirtai), meaning “rope-fasteners”.

7

o. Introduction

((Construct the tine perpendicular to t passing through P.})
RightAngle( l , P ):

Choose a point

A,B <— Intersect(Circle(P,A), t)

C,D <— Intersect(Circle(A,B), Circle(B,A))
return Line(C, D)

({Construct a point Z suchthat \AZ\ = |AC||AD|/|AB|.))
MultiplyOrDivide(A, B, C, D ):
a <— RightAngle(Line(A, C),A)

E <— Intersect(Circle(A,B), a)

F <— Intersect(Circle(A,D), a)

/5 <— RightAngle(Line(£, C),F )

Y <— RightAngle(/ 3,.F)
return Intersect(t, Line(A, C))

Figure 0.4. Multiplication by compass and straightedge.

similar. The second and third lines of the main algorithm are ambiguous,
because a intersects any circle centered at A at two distinct points, but the
algorithm is actually correct no matter which intersection points are chosen for
E and F.

Euclid’s algorithm reduces the problem of multiplying two magnitudes
(lengths) to a series of primitive compass-and-straightedge operations. These
operations are difficult to implement precisely on a modern digital computer, but
Euclid’s algorithm wasn’t designed for a digital computer. It was designed for the
Platonic Ideal Geometer, wielding the Platonic Ideal Compass and the Platonic
Ideal Straightedge, who could execute each operation perfectly in constant time
by definition. In this model of computation, MultiplyOrDivide runs in 0(1)
time!

0.3 Congressional Apportionment

Here is another real-world example of an algorithm of significant political
importance. Article I, Section 2 of the United States Constitution requires that

Representatives and direct Taxes shall be apportioned among the several
States which may be included within this Union, according to their respective
Numbers.... The Number of Representatives shall not exceed one for every
thirty Thousand, but each State shall have at Least one Representative....

Because there are only a finite number of seats in the House of Representatives,
exact proportional representation requires either shared or fractional represen¬
tatives, neither of which are legal. As a result, over the next several decades,
many different apportionment algorithms were proposed and used to round
the ideal fractional solution fairly. The algorithm actually used today, called
the Huntington-Hill method or the method of equal proportions, was first

8

0.3- Congressional Apportionment

suggested by Census Bureau statistician Joseph Hill in 1911, refined by Harvard
mathematician Edward Huntington in 1920, adopted into Federal law (2 U.S.C.
§2a) in 1941, and survived a Supreme Court challenge in 1992. 13

The Huntington-Hill method allocates representatives to states one at a
time. First, in a preprocessing stage, each state is allocated one representative.
Then in each iteration of the main loop, the next representative is assigned
to the state with the highest priority. The priority of each state is defined
to be P /*j r(r + 1 ), where P is the state’s population and r is the number of
representatives already allocated to that state.

The algorithm is described in pseudocode in Figure 0.5. The input consists
of an array Pop[l .. n] storing the populations of the n states and an integer if
equal to the total number of representatives. (Currently, in the United States,
n = 50 and R = 435 .) The output array Rep[ 1 .. n] records the number of
representatives allocated to each state.

APPORTIQNCONGRESSfPopf 1.. n],R):

PQ <— NewPriorityQueue

for i <— 1 to n
Rep[i] <— 1

Insert (PQ, i, Pop[i]/V 2)

P<-P- 1

for i «— n + 1 to R

s <— ExtractMax(PQ)

Pep[s] <— Pep[s] + 1

priority <— Pop [ 5 ] j V^PepflJfPepfJJTTj

Insert {PQ,s,priority)

return Rep[ 1.. n]

Figure 0.5. The Huntington-Hill apportionment algorithm

This implementation of Huntington-Hill uses a priority queue that supports
the operations NewPriorityQueue, Insert, and ExtractMax. (The actual
law doesn’t say anything about priority queues, of course.) The output of the
algorithm, and therefore its correctness, does not depend at all on how this
priority queue is implemented. The Census Bureau uses a sorted array, stored
in a single column of an Excel spreadsheet, which is recalculated from scratch

13 Overruling an earlier ruling by a federal district court, the Supreme Court unanimously
held that any apportionment method adopted in good faith by Congress is constitutional (United
States Department of Commerce v. Montana ). The current congressional apportionment algorithm
is described in gruesome detail at the U.S. Census Department web site http://www.census.gov/
topics/public-sector/congressional-apportionment.html. A good history of the apportionment
problem can be found at http://www.thirty-thousand.org/pages/Apportionment.htm. A report
by the Congressional Research Service describing various apportionment methods is available at
http://www.fas.org/sgp/crs/misc/R41382.pdf.

o. Introduction

at every iteration. You (should have) learned a more efficient implementation
in your undergraduate data structures class.

Similar apportionment algorithms are used in multi-party parliamentary
elections around the world, where the number of seats allocated to each party
is supposed to be proportional to the number of votes that party receives. The
two most common are the D’Hondt method 14 and the Webster-Sainte-Lague
method , 15 which respectively use priorities P/(r + 1 ) and P/( 2 r + 1 ) in place of
the square-root expression in Huntington-Hill. The Huntington-Hill method is
essentially unique to the United States House of Representatives, thanks in part
to the constitutional requirement that each state must be allocated at least one
representative.

0.4 A Bad Example

As a prototypical example of a sequence of instructions that is not actually an
algorithm, consider "Martin’s algorithm”: 16

BeAMillionaireAndNeverPayTaxesQ:

Get a million dollars.

If the tax man comes to your door and says, “You have never paid taxes!"
Say “/ forgot."

Pretty simple, except for that first step; it’s a doozy. A group of billionaire CEOs,
Silicon Valley venture capitalists, or New York City real-estate hustlers might
consider this an algorithm, since for them the first step is both unambiguous
and trivial, 17 but for the rest of us poor slobs, Martin’s procedure is too vague to
be considered an actual algorithm. On the other hand, this is a perfect example
of a reduction —it reduces the problem of being a millionaire and never paying
taxes to the “easier” problem of acquiring a million dollars. We’ll see reductions
over and over again in this class. As hundreds of businessmen and politicians
have demonstrated, if you know how to solve the easier problem, a reduction
tells you how to solve the harder one.

Martin’s algorithm, like some of our previous examples, is not the kind
of algorithm that computer scientists are used to thinking about, because it

14 developed by Thomas Jefferson in 1792, used for U.S. Congressional apportionment from
1792 to 1832, rediscovered by Belgian mathematician Victor D’Hondt in 1878, and refined by Swiss
physicist Eduard Hagenbach-Bischoff in 1888.

15 developed by Daniel Webster in 1832, used for U.S. Congressional apportionment from 1842
to 1911, rediscovered by French mathematician Andre Sainte-Lague in 1910, and rediscovered
again by German physicist Hans Schepers in 1980.

l6 Steve Martin, “You Can Be A Millionaire”, Saturday Night Live, January 21, 1978. Also
appears on Comedy Is Not Pretty, Warner Bros. Records, 1979.

17 Something something secure quantum blockchain deep-learning something.

10

0.5. Describing Algorithms

is phrased in terms of operations that are difficult for computers to perform.
This book focuses (almost!) exclusively on algorithms that can be reasonably
implemented on a standard digital computer. Each step in these algorithms
is either directly supported by common programming languages (such as
arithmetic, assignments, loops, or recursion) or something that you’ve already
learned how to do (like sorting, binary search, tree traversal, or singing “n
Bottles of Beer on the Wall”).

0.5 Describing Algorithms

The skills required to effectively design and analyze algorithms are entangled
with the skills required to effectively describe algorithms. At least in my classes,
a complete description of any algorithm has four components:

• What: A precise specification of the problem that the algorithm solves.

• How: A precise description of the algorithm itself.

• Why: A proof that the algorithm solves the problem it is supposed to solve.

• How fast: An analysis of the running time of the algorithm.

It is not necessary (or even advisable) to develop these four components in this
particular order. Problem specifications, algorithm descriptions, correctness
proofs, and time analyses usually evolve simultaneously, with the development
of each component informing the development of the others. For example,
we may need to tweak the problem description to support a faster algorithm,
or modify the algorithm to handle a tricky case in the proof of correctness.
Nevertheless, presenting these components separately is usually clearest for the
reader.

As with any writing, it’s important to aim your descriptions at the right
audience; I recommend writing for a competent but skeptical programmer who
is not as clever as you are. Think of yourself six months ago. As you develop any
new algorithm, you will naturally build up lots of intuition about the problem
and about how your algorithm solves it, and your informal reasoning will be
guided by that intuition. But anyone reading your algorithm later, or the code
you derive from it, won’t share your intuition or experience. Neither will your
compiler. Neither will you six months from now. All they will have is your
written description.

Even if you never have to explain your algorithms to anyone else, it’s still
important to develop them with an audience in mind. Trying to communicate
clearly forces you to think more clearly. In particular, writing for a novice
audience, who will interpret your words exactly as written, forces you to work
through fine details, no matter how “obvious” or “intuitive” your high-level ideas
may seem at the moment. Similarly, writing for a skeptical audience forces you

11

o. Introduction

to develop robust arguments for correctness and efficiency, instead of trusting
your intuition or your intelligence. 18

I cannot emphasize this point enough: Your primary job as an algorithm
designer is teaching other people how and why your algorithms work. If

you can’t communicate your ideas to other human beings, they may as well
not exist. Producing correct and efficient executable code is an important
but secondary goal. Convincing yourself, your professors, your (prospective)
employers, your colleagues, or your students that you are smart is at best a
distant third.

Specifying the Problem

Before we can even start developing a new algorithm, we have to agree on what
problem our algorithm is supposed to solve. Similarly, before we can even start
describing an algorithm, we have to describe the problem that the algorithm is
supposed to solve.

Algorithmic problems are often presented using standard English, in terms of
real-world objects, not in terms of formal mathematical objects. It’s up to us, the
algorithm designers, to restate these problems in terms of abstract mathematical
objects—numbers, arrays, lists, graphs, trees, and so on—that we can reason
about formally. We must also determine if the problem statement carries any
hidden assumptions, and state those assumptions explicitly. (For example, in
the song “n Bottles of Beer on the Wall”, n is always a non-negative integer. 19 )

We may need to refine our specification as we develop the algorithm. For
example, our algorithm may require a particular input representation, or
produce a particular output representation, that was left unspecified in the
original informal problem description. Or our algorithm might actually solve a
more general problem than we were originally asked to solve. (This is a common
feature of recursive algorithms.)

The specification should include just enough detail that someone else could
use our algorithm as a black box, without knowing how or why the algorithm
actually works. In particular, we must describe the type and meaning of each
input parameter, and exactly how the eventual output depends on the input
parameters. On the other hand, our specification should deliberately hide any
details that are not necessary to use the algorithm as a black box. Let that which
does not matter truly slide.

For example, the lattice and duplation-and-mediation algorithms both solve
the same problem: Given two non-negative integers x and y, each represented

l8 In particular, I assume that you are a skeptical novice!

19 I’ve never heard anyone sing “-/2 Bottles of Beer on the Wall.” Occasionally I have heard set
theorists singing “H 0 bottles of beer on the wall”, but for some reason they always gave up before
the song was over.

12

0.5. Describing Algorithms

as an array of digits, compute the product x • y, also represented as an array of
digits. To someone using these algorithms, the choice of algorithm is completely
irrelevant. On the other hand, the Greek straightedge-and-compass algorithm
solves a different problem, because the input and output values are represented
by line segments instead of arrays of digits.

Describing the Algorithm

Computer programs are concrete representations of algorithms, but algorithms
are not programs. Rather, algorithms are abstract mechanical procedures
that can be implemented in any programming language that supports the
underlying primitive operations. The idiosyncratic syntactic details of your
favorite programming language are utterly irrelevant; focusing on these will
only distract you (and your readers) from what’s really going on. 20 A good
algorithm description is closer to what we should write in the comments of a
real program than the code itself. Code is a poor medium for storytelling.

On the other hand, a plain English prose description is usually not a good idea
either. Algorithms have lots of idiomatic structure—especially conditionals, loops,
function calls, and recursion—that are far too easily hidden by unstructured
prose. Colloquial English is full of ambiguities and shades of meaning, but
algorithms must be described as unambiguously as possible. Prose is a poor
medium for precision.

In my opinion, the clearest way to present an algorithm is using a combination
of pseudocode and structured English. Pseudocode uses the structure of formal
programming languages and mathematics to break algorithms into primitive
steps; the primitive steps themselves can be written using mathematical notation,
pure English, or an appropriate mixture of the two, whatever is clearest. Well-
written pseudocode reveals the internal structure of the algorithm but hides
irrelevant implementation details, making the algorithm easier to understand,
analyze, debug, and implement.

Whenever we describe an algorithm, our description should include every
detail necessary to fully specify the algorithm, prove its correctness, and analyze

“This is, of course, a matter of religious conviction. Linguists argue incessantly over the
Sapir-Whorf hypothesis, which states (more or less) that people think only in the categories
imposed by their languages. According to an extreme formulation of this principle, some concepts
in one language simply cannot be understood by speakers of other languages, not just because of
technological advancement—How would you translate “jump the shark” or “Fortnite streamer”
into Aramaic?—but because of inherent structural differences between languages and cultures.
For a more skeptical view, see Steven Pinker’s The Language Instinct. There is admittedly some
strength to this idea when applied to different programming paradigms. (What’s the Y combinator,
again? How do templates work? What’s an Abstract Factory?) Fortunately, those differences are
too subtle to have any impact on the material in this book. For a compelling counterexample, see
Chris Okasaki’s thesis/monograph Functional Data Structures and its more recent descendants.

13

o. Introduction

its running time. At the same time, it should exclude any details that are not
necessary to fully specify the algorithm, prove its correctness, and analyze its
running time. (Slide.) At a more practical level, our description should allow
a competent but skeptical programmer who has not read this book to quickly
and correctly implement the algorithm in their favorite programming language,
without understanding why it works.

I don’t want to bore you with the rules I follow for writing pseudocode, but
I must caution against one especially pernicious habit. Never describe repeated
operations informally, as in “Do [this] first, then do [that] second, and so on." or
“Repeat this process until [something]”. As anyone who has taken one of those
frustrating “What comes next in this sequence?” tests already knows, describing
the first few steps of an algorithm says little or nothing about what happens
in later steps. If your algorithm has a loop, write it as a loop, and explicitly
describe what happens in an arbitrary iteration. Similarly, if your algorithm is
recursive, write it recursively, and explicitly describe the case boundaries and
what happens in each case.

0.6 Analyzing Algorithms

It’s not enough just to write down an algorithm and say “Behold!” We must also
convince our audience (and ourselves!) that the algorithm actually does what
it’s supposed to do, and that it does so efficiently.

Correctness

In some application settings, it is acceptable for programs to behave correctly
most of the time, on all “reasonable” inputs. Not in this book; we require
algorithms that are always correct, for all possible inputs. Moreover, we must
prove that our algorithms are correct; trusting our instincts, or trying a few test
cases, isn’t good enough. Sometimes correctness is truly obvious, especially
for algorithms you’ve seen in earlier courses. On the other hand, “obvious”
is all too often a synonym for “wrong”. Most of the algorithms we discuss in
this course require real work to prove correct. In particular, correctness proofs
usually involve induction. We like induction. Induction is our friend . 21

Of course, before we can formally prove that our algorithm does what it’s
supposed to do, we have to formally describe what it’s supposed to do!

Running Time

The most common way of ranking different algorithms for the same problem is
by how quickly they run. Ideally, we want the fastest possible algorithm for any

21 If induction is not your friend, you will have a hard time with this book.

14

0 . 6 . Analyzing Algorithms

particular problem. In many application settings, it is acceptable for programs
to run efficiently most of the time, on all “reasonable” inputs. Not in this class;
we require algorithms that always run efficiently, even in the worst case.

But how do we measure running time? As a specific example, how long does
it take to sing the song BottlesOfBeer(u)? This is obviously a function of the
input value n, but it also depends on how quickly you can sing. Some singers
might take ten seconds to sing a verse; others might take twenty. Technology
widens the possibilities even further. Dictating the song over a telegraph using
Morse code might take a full minute per verse. Downloading an mp3 over
the Web might take a tenth of a second per verse. Duplicating the mp3 in a
computer’s main memory might take only a few microseconds per verse.

What’s important here is how the singing time changes as n grows. Singing
BoTTLEsOFBEER(2n) requires about twice much time as singing BottlesOf-
Beer(u), no matter what technology is being used. This is reflected in the
asymptotic singing time 0(n).

We can measure time by counting how many times the algorithm executes a
certain instruction or reaches a certain milestone in the “code”. For example,
we might notice that the word “beer” is sung three times in every verse of
BottlesOfBeer, so the number of times you sing “beer” is a good indication
of the total singing time. For this question, we can give an exact answer:
BottlesOfBeer(u) mentions beer exactly 3n + 3 times.

Incidentally, there are lots of other songs with quadratic singing time. This
one is probably familiar to most English-speakers:

NDAYSOFCHRISTMAS(gi/tsr2 .. n]):

for i <— 1 to n

Sing “On the ith day of Christmas, my true love gave to me”
for j <— i down to 2
Sing “j gifts[j ]”
if i> 1

Sing “and”

Sing “a partridge in a pear tree.”

The input to NDaysOfChristmas is a list of n — 1 gifts, represented here as
an array. It’s quite easy to show that the singing time is 0 (n 2 ); in particular,
the singer mentions the name of a gift ^"=1 i = n(n + l)/2 times (counting the
partridge in the pear tree). It’s also easy to see that during the first n days of
Christmas, my true love gave to me exactly 2j = i J = n ( n + l)( n + 2)/6 =
0(n 3 ) gifts.

Other quadratic-time songs include “Old MacDonald Had a Farm”, “There
Was an Old Lady Who Swallowed a Fly”, “Hole in the Bottom of the Sea”, “Green
Grow the Rushes O”, “The Rattlin’ Bog”, “The Court Of King Caractacus”,“The
Barley-Mow”, “If I Were Not Upon the Stage”, “Star Trekkin’ ”, “ 1 st das nicht

15

o. Introduction

ein Schnitzelbank?”, 22 “Il Pulcino Pio”, “Minkurinn i h as ns n akof a n u m”, “Echad
Mi Yodea”, and “To kokoquxi”. For more examples, consult your favorite
preschooler.

ALOUETTE(Zapartl 1.. nl):

Chantez «Alouette, gentille alouette, alouette, je te plumerais. »

pour tout i de 1 a n

Chantez « Je te plumerais lapart[i ]. Je te plumerais lapart[i], »
pour tout j de i — 1 a bas a 1

Chantez « Et lapart[j ] I Et lapart[j ] ! »

Chantez « Alouette, alouette, ooooooo!»

Chantez « Alouette, gentille allouette, alouette, je te plumerais. »

A few songs have even more bizarre singing times. A fairly modern example
is “The TELNET Song” by Guy Steele, which actually takes 0 ( 2 n ) time to sing
the first n verses; Steele recommended n = 4 . Finally, there are some songs that
never end. 23

Except for “The TELNET Song”, all of these songs are iterative algorithms,
expressed as a small set of nested loops, so their running singing times can be
computed using nested summations. The running time of a recursive algorithm is
more easily expressed as a recurrence. For example, the peasant multiplication
algorithm can be expressed recursively as follows:

{ 0 if x = 0

Lx/2j-(y + y) if x is even
[x/2j-(y + y) + y ifxisodd

Let T(x,y) denote the number of parity, addition, and mediation operations
required to compute x • y. This function satisfies the recursive inequality
T(x,y) < T([x/ 2 J, 2 y) + 2 with base case T( 0 ,y) = 0 . Techniques described
in the next chapter imply the upper bound T(x,y) = O(logx).

Sometimes the running time of an algorithm depends on a particular
implementation of some underlying data structure of subroutine. For example,
the Huntington-Hill apportionment algorithm ApportionCongress runs in
0(1V +RI + ( R — n ) E ) time, where N denotes the running time of NewPriority-
Queue, I denotes the running time of Insert, and E denotes the running time
of ExtractMax. Under the reasonable assumption that R > 2 n (on average,
each state gets at least two representatives), we can simplify this bound to
0 (N +R(I + £)). The precise running time depends on the implementation
of the underlying priority queue. The Census Bureau implements the priority

22 Ja, das ist Otto von Schnitzelpusskrankengescheitmeyer!

23 They just go on and on, my friend.

16

Exercises

queue as an unsorted array, which gives us N = I = 0(1) and E = 0(n), so the
Census Bureau’s implementation of ApportionCongress runs in 0(Rn) time.
However, if we implement the priority queue as a binary heap or a heap-ordered
array, we have N = 0(1) and I =R = 0(log n), so the overall algorithm runs in
O(Rlogn) time.

Finally, sometimes we are interested in computational resources other than
time, such as space, number of coin flips, number of cache or page faults, number
of inter-process messages, or the number of gifts my true love gave to me. These
resources can be analyzed using the same techniques used to analyze running
time. For example, lattice multiplication of two n-digit numbers requires 0(n 2 )
space if we write down all the partial products before adding them, but only
O(n) space if we add them on the fly.

Exercises

o. Describe and analyze an efficient algorithm that determines, given a legal
arrangement of standard pieces on a standard chess board, which player will
win at chess from the given starting position if both players play perfectly.
[Hint: There is a trivial one-line solution!]

v i. (a) Identify (or write) a song that requires 0(n 3 ) time to sing the first n
verses.

(b) Identify (or write) a song that requires 0(nlogn) time to sing the first
n verses.

(c) Identify (or write) a song that requires some other weird amount of
time to sing the first n verses.

2 . Careful readers might complain that our analysis of songs like “n Bottles of
Beer on the Wall” or “The n Days of Christmas” is overly simplistic, because
larger numbers take longer to sing than shorter numbers. More generally,
because there are only so many words of a given length, larger sets of words
necessarily contain longer words. 24 We can more accurately estimate singing
time by counting the number of syllables sung, rather than the number of
words.

(a) How long does it take to sing the integer n?

(b) How long does it take to sing “n Bottles of Beer on the Wall”?

(c) How long does it take to sing “The n Days of Christmas”?

As usual, express your answers in the form 0(/(n)) for some function /.

24 Ja, das ist das Subatomarteilchenbeschleunigungsnaturmafiigkeitsuntersuchungsmaschine!

17

o. Introduction

3 . The cumulative drinking song “The Barley Mow” has been sung throughout
the British Isles for centuries. The song has many variants; Figure 0.6
contains pseudolyrics for one version traditionally sung in Devon and
Cornwall, where vessel[i] is the name of a vessel that holds 2 l ounces of
beer . 25

BarleyMowCh):

"Here's a health to the barley-mow, my brave boys,"
"Here's a health to the barley-mow!"

"We'll drink it out of the jolly brown bowl,"

"Here's a health to the barley-mow!"

"Here's a health to the barley-mow, my brave boys,"
"Here's a health to the barley-mow!"

for i <— 1 to n

"We’ll drink it out of the vessel[i], boys,"

"Here's a health to the barley-mowI"
for j <— i downto 1
"The vessel[j ],"

"And the jolly brown bowl!"

"Here's a health to the barley-mow!"

"Here's a health to the barley-mow, my brave boys,"
"Here's a health to the barley-mow!"

Figure 0.6. "The Barley Mow".

(a) Suppose each name vessel[i ] is a single word, and you can sing four
words a second. How long would it take you to sing BarleyMow(u)?
(Give a tight asymptotic bound.)

(b) If you want to sing this song for arbitrarily large values of n, you’ll have
to make up your own vessel names. To avoid repetition, these names
must become progressively longer as rt increases. Suppose vessel[n ] has
0(log rt) syllables, and you can sing six syllables per second. Now how
long would it take you to sing BarleyMow(u)? (Give a tight asymptotic
bound.)

25 In practice, the song uses some subset of the following vessels; nipperkin, quarter-gill,
half-a-gill, gill, quarter-pint, half-a-pint, pint, quart, pottle, gallon, half-anker, anker, firkin,
half-barrel/kilderkin, barrel, hogshead, pipe/butt, tun, well, river, and ocean. With a few
exceptions (especially at the end), every vessel in this list has twice the volume of its predecessor.
Irish and Scottish versions of the song have slightly different lyrics, and they usually switch to
people (barmaid, landlord, drayer, and so on) after “gallon”.

An early version of the song entitled “Give us once a drink” appears in the play Jack Drum’s
Entertainment (or the Comedie ofPasquill and Katherine) written by John Marston around 1600.
(“Giue vs once a drinke for and the black bole. Sing gentle Butler bally moy\”) There is some
disagreement whether Marston wrote the “high Dutch Song” specifically for the play, whether
“bally moy” is a mondegreen for “barley mow” or vice versa, or whether it’s actually the same
song at all. These discussions are best had over n bottles of beer.

18

Exercises

(c) Suppose each time you mention the name of a vessel, you actually drink
the corresponding amount of beer: one ounce for the jolly brown bowl,
and 2 ! ounces for each vessel[i]. Assuming for purposes of this problem
that you are at least 21 years old, exactly how many ounces of beer would
you drink if you sang BarleyMow(u)? (Give an exact answer, not just
an asymptotic bound.)

4. Recall that the input to the Huntington-Hill algorithm ApportionCongress
is an array P[1.. n ], where P[i] is the population of the ith state, and an
integer R, the total number of representatives to be allotted. The output is
an array r[l.. n], where r[i] is the number of representatives allotted to the
ith state by the algorithm.

Let P = P[i] denote the total population of the country, and let
r* = R - P\i]/P denote the ideal number of representatives for the ith state.

(a) Prove that r[i] > [r*J for all i.

(b) Describe and analyze an algorithm that computes exactly the same
congressional apportionment as ApportionCongress in 0(n log n) time.
(Recall that the running time of ApportionCongress depends on R,
which could be arbitrarily larger than n.)

r (c) If a state’s population is small relative to the other states, its ideal number
r* of representatives could be close to zero; thus, tiny states are over¬
represented by the Huntington-Hill apportionment process. Surprisingly,
this can also be true of very large states. Let a = (1 + V 2)/2 & 1.20711.
Prove that for any e > 0, there is an input to ApportionCongress with
maXj P[i] = P[l], such that r[ 1 ] > (a — e) rj\

* (d) Can you improve the constant a in the previous question?

The control of a large force Is the same principle as the control of a few men:

It Is merely a question of dividing up their numbers.

- Sun Zi, The Art of War ( c .400 ce ), translated by Lionel Giles (igio)

Our life is frittered away by detail.... Simplify, simplify.

— Henry David Thoreau, Walden (1854)

Now, don't ask me what Voom is. I never will know.

But, boy! Let me tell you, it DOES clean up snow!

— Dr. Seuss [Theodor Seuss Geisel], The Cat in the Hat Comes Back (1958)

Do the hard jobs first. The easy jobs will take care of themselves.

— attributed to Dale Carnegie

Recursion

1.1 Reductions

Reduction is the single most common technique used in designing algorithms.
Reducing one problem X to another problem Y means to write an algorithm
forX that uses an algorithm for Y as a black box or subroutine. Crucially, the
correctness of the resulting algorithm for X cannot depend in any way on how
the algorithm for Y works. The only thing we can assume is that the black box
solves Y correctly. The inner workings of the black box are simply none of our
business; they’re somebody else’s problem. It’s often best to literally think of the
black box as functioning purely by magic.

For example, the peasant multiplication algorithm described in the previous
chapter reduces the problem of multiplying two arbitrary positive integers to
three simpler problems: addition, mediation (halving), and parity-checking. The
algorithm relies on an abstract “positive integer” data type that supports those
three operations, but the correctness of the multiplication algorithm does not

21

i. Recursion

depend on the precise data representation (tally marks, clay tokens, Babylonian
hexagesimal, quipu, counting rods, Roman numerals, finger positions, augrym
stones, gobar numerals, binary, negabinary, Gray code, balanced ternary, phinary,
quater-imaginary, ... ), or on the precise implementations of those operations.
Of course, the running time of the multiplication algorithm depends on the
running time of the addition, mediation, and parity operations, but that’s
a separate issue from correctness. Most importantly, we can create a more
efficient multiplication algorithm just by switching to a more efficient number
representation (from tally marks to place-value notation, for example).

Similarly, the Huntington-Hill algorithm reduces the problem of apportioning
Congress to the problem of maintaining a priority queue that supports the
operations Insert and ExtractMax. The abstract data type “priority queue” is
a black box; the correctness of the apportionment algorithm does not depend
on any specific priority queue data structure. Of course, the running time of
the apportionment algorithm depends on the running time of the Insert and
ExtractMax algorithms, but that’s a separate issue from the correctness of the
algorithm. The beauty of the reduction is that we can create a more efficient
apportionment algorithm by simply swapping in a new priority queue data
structure. Moreover, the designer of that data structure does not need to know
or care that it will be used to apportion Congress.

When we design algorithms, we may not know exactly how the basic building
blocks we use are implemented, or how our algorithms might be used as building
blocks to solve even bigger problems. That ignorance is uncomfortable for many
beginners, but it is both unavoidable and extremely useful. Even when you
do know precisely how your components work, it is often extremely helpful to
pretend that you don’t.

i .2 Simplify and Delegate

Recursion is a particularly powerful kind of reduction, which can be described
loosely as follows:

• If the given instance of the problem can be solved directly, solve it directly.

• Otherwise, reduce it to one or more simpler instances of the same problem.

If the self-reference is confusing, it may be helpful to imagine that someone else
is going to solve the simpler problems, just as you would assume for other types
of reductions. I like to call that someone else the Recursion Fairy. Your only
task is to simplify the original problem, or to solve it directly when simplification
is either unnecessary or impossible; the Recursion Fairy will solve all the simpler
subproblems for you, using Methods That Are None Of Your Business So Butt

22

1.2. Simplify and Delegate

Out. 1 Mathematically sophisticated readers might recognize the Recursion Fairy
by its more formal name: the Induction Hypothesis.

There is one mild technical condition that must be satisfied in order for
any recursive method to work correctly: There must be no infinite sequence of
reductions to simpler and simpler instances. Eventually, the recursive reductions
must lead to an elementary base case that can be solved by some other method;
otherwise, the recursive algorithm will loop forever. The most common way
to satisfy this condition is to reduce to one or more smaller instances of the
same problem. For example, if the original input is a skreeble with n glurps, the
input to each recursive call should be a skreeble with strictly less than n glurps.
Of course this is impossible if the skreeble has no glurps at all—You can’t have
negative glurps; that would be silly!—so in that case we must grindlebloff the
skreeble using some other method.

We’ve already seen one instance of this pattern in the peasant multiplication
algorithm, which is based directly on the following identity.

[ 0 if x = 0

x ~y = t L*/2J • (y + y) if x is even

[l*/2 J • (y + y) + y if x is odd

The same identity can be expressed algorithmically as follows:

Multiply(x, y):

if x = 0

return 0

else

x ' <- |x/2j

y'<-y + y

prod <— Multiplex', y') ((Recurse!))
if x is odd

prod <— prod + y
return prod

A lazy Egyptian scribe could execute this algorithm by computing x' and y\
asking a more junior scribe to multiply x' and y' , and then possibly adding y
to the junior scribe’s response. The junior scribe’s problem is simpler because
x 7 < x, and repeatedly decreasing a positive integer eventually leads to 0. How
the junior scribe actually computes x 7 • y' is none of the senior scribe’s business
(and it’s none of your business, either).

‘When I was an undergraduate, I attributed recursion to “elves” instead of the Recursion Fairy,
referring to the Brothers Grimm story about an old shoemaker who leaves his work unfinished
when he goes to bed, only to discover upon waking that elves (“Wichtelmanner”) have finished
everything overnight. Someone more entheogenically experienced than I might recognize these
Rekursionswichtelmanner as Terence McKenna’s “self-transforming machine elves”.

23

i. Recursion

i .3 Tower of Hanoi

The Tower of Hanoi puzzle was first published—as an actual physical puzzle!—by
the French teacher and recreational mathematician Eduoard Lucas in r 883, 2

under the pseudonym “N. Claus (de Siam)” (an anagram of “Lucas d’Amiens”).

The following year, Henri de Parville described the puzzle with the following
remarkable story : 3

In the great temple at Benares 4 ... beneath the dome which marks the centre of
the world, rests a brass plate in which are fixed three diamond needles, each
a cubit high and as thick as the body of a bee. On one of these needles, at the
creation, God placed sixty-four discs of pure gold, the largest disc resting on the
brass plate, and the others getting smaller and smaller up to the top one. This is
the Tower of Bramah. Day and night unceasingly the priests transfer the discs
from one diamond needle to another according to the fixed and immutable
laws of Bramah, which require that the priest on duty must not move more
than one disc at a time and that he must place this disc on a needle so that
there is no smaller disc below it. When the sixty-four discs shall have been thus
transferred from the needle on which at the creation God placed them to one
of the other needles, tower, temple, and Brahmins alike will crumble into dust,
and with a thunderclap the world will vanish.

Figure 1.1. The Tower of Hanoi puzzle

Of course, as good computer scientists, our first instinct on reading this
story is to substitute the variable n for the hardwired constant 64. And because
most physical instances of the puzzle are made of wood instead of diamonds
and gold, I will call the three possible locations for the disks “pegs” instead of

2 Lucas later claimed to have invented the puzzle in 1876.

3 This English translation is taken from W. W. Rouse Ball’s 1892 book Mathematical Recreations
and Essays.

4 The “great temple at Benares” is almost certainly the Kashi Vishvanath Temple in Varanasi,
Uttar Pradesh, India, located approximately 2400km west-north-west of Ha Noi, Viet Nam, where
the fictional N. Claus supposedly resided. Coincidentally, the French Army invaded Hanoi in 1883,
the same year Lucas released his puzzle, ultimately leading to its establishment as the capital of
French Indochina.

24

1-3- Tower of Hanoi

“needles”. How can we move a tower of n disks from one peg to another, using a
third peg as an occasional placeholder, without ever placing a disk on top of a
smaller disk?

As N. Claus (de Siam) pointed out in the pamphlet included with his puzzle,
the secret to solving this puzzle is to think recursively. Instead of trying to solve
the entire puzzle at once, let’s concentrate on moving just the largest disk. We
can’t move it at the beginning, because all the other disks are in the way. So
first we have to move those n — 1 disks to the third peg. And then after we move
the largest disk, we have to move those n — 1 disks back on top of it.

1

| recursion j>

J

h Z 1 —m

1

Jr

Figure i .2. The Tower of Hanoi algorithm; ignore everything but the bottom disk.

So now all we have to figure out is how to—

NO!! STOP!!

That’s it! We’re done! We’ve successfully reduced the n-disk Tower of Hanoi
problem to two instances of the (n — l)-disk Tower of Hanoi problem, which
we can gleefully hand off to the Recursion Fairy—or to carry Lucas’s metaphor
further, to the junior monks at the temple. Our job is finished. If we didn’t trust
the junior monks, we wouldn’t have hired them; let them do their job in peace.

Our reduction does make one subtle but extremely important assumption:
There is a largest disk. Our recursive algorithm works for any positive number
of disks, but it breaks down when n = 0. We must handle that case using a
different method. Fortunately, the monks at Benares, being good Buddhists, are
quite adept at moving zero disks from one peg to another in no time at all, by
doing nothing.

Figure i .3. The vacuous base case for the Tower of Hanoi algorithm. There is no spoon.

It may be tempting to think about how all those smaller disks move around—
or more generally, what happens when the recursion is unrolled—but really,
don’t do it. For most recursive algorithms, unrolling the recursion is neither

25

i. Recursion

necessary nor helpful. Our only task is to reduce the problem instance we’re
given to one or more simpler instances, or to solve the problem directly if such
a reduction is impossible. Our recursive Tower of Hanoi algorithm is trivially
correct when n = 0 . For any n > 1 , the Recursion Fairy correctly moves the top
n — 1 disks (more formally, the Inductive Hypothesis implies that our recursive
algorithm correctly moves the top n — 1 disks) so our algorithm is correct.

The recursive Hanoi algorithm is expressed in pseudocode in Figure 1.4.
The algorithm moves a stack of n disks from a source peg (src) to a destination
peg (dst) using a third temporary peg (tmp) as a placeholder. Notice that the
algorithm correctly does nothing at all when n = 0 .

Hanoi(h, src, dst, tmp):

if n > 0

Hanoi(h — l,src, tmp, dst )
move disk n from src to dst

((Recurse!))

Hanoi(h — 1, tmp, dst,src)

((Recurse!))

Figure 1.4. A recursive algorithm to solve the Tower of Hanoi

Let T(n ) denote the number of moves required to transfer n disks—the
running time of our algorithm. Our vacuous base case implies that T( 0 ) = 0 ,
and the more general recursive algorithm implies that T(n) = 2 T(n — 1 ) + 1
for any n > 1 . By writing out the first several values of T(n), we can easily
guess that T(n) = 2' ! — 1; a straightforward induction proof implies that this
guess is correct. In particular, moving a tower of 64 disks requires 2 64 — 1 =
r8,446,744,073,709,551,6t5 individual moves. Thus, even at the impressive rate
of one move per second, the monks at Benares will be at work for approximately
585 billion years (“plus de cinq millards de siecles”) before tower, temple, and
Brahmins alike will crumble into dust, and with a thunderclap the world will
vanish.

1.4 Mergesort

Mergesort is one of the earliest sorting algorithms designed for general-purpose
stored-program computers. The algorithm was developed by John von Neumann
in 1945, and described in detail in a publication with Herman Goldstine in t947,
as one of the first non-numerical programs for the EDVAC. 5

1. Divide the input array into two subarrays of roughly equal size.

5 Goldstine and von Neumann actually described an non-recursive variant now usually called
bottom-up mergesort. At the time, large data sets were sorted by special-purpose machines—
almost all built by IBM—that manipulated punched cards using variants of binary radix sort.
Von Neumann argued (successfully!) that because the EDVAC could sort faster IBM’s dedicated
sorters, “without human intervention or need for additional equipment”, the EDVAC was an “all
purpose” machine, and special-purpose sorting machines were no longer necessary.

26

14 - Mergesort

2. Recursively mergesort each of the subarrays.

3. Merge the newly-sorted subarrays into a single sorted array.

Input:

S

0

R

T

I

N

G

E

X

A

M

P

L

Divide:

S

0

R

T

I

N

G

E

X

A

M

P

L

Recurse Left:

I

N

0

S

R

T

G

E

X

A

M

P

L

Recurse Right:

I

N

0

S

R

T

A

E

G

L

M

P

X

Merge:

A

E

G

I

L

M

N

0

P

R

S

T

X

Figure 1.5. A mergesort example.

The first step is completely trivial—-just divide the array size by two—and
we can delegate the second step to the Recursion Fairy. All the real work is
done in the final merge step. A complete description of the algorithm is given in
Figure 1.6; to keep the recursive structure clear, I’ve extracted the merge step
into an independent subroutine. The merge algorithm is also recursive—identify
the first element of the output array, and then recursively merge the rest of the
input arrays.

Figure 1.6. Mergesort

MergeSort(A[ 1 .. nl):

if n > 1

m <— [n/2\

MergeSort(A [1 .. m]) ((Recurse!))
MergeSort(A[/r + 1 .. n]) ((Recurse!))
Merge(A [1 ..n],m)

Merge(A[ 1.. nl, m)\

i <— 1; j m + 1
for k «— 1 to n

if j > n

B[k]

-A[i] ;

i <—

i +1

else if i >

m

B[k ]

«- A[j ];

J + l

else if A[i

]<A[j]

B[k ]

-A[i];

i <—

i + 1

else

B[k]

-4 j];

J + l

: <— 1 to n

A[fc]«-B[fc]

Correctness

To prove that this algorithm is correct, we apply our old friend induction twice,
first to the Merge subroutine then to the top-level Mergesort algorithm.

Lemma 1.1. Merge correctly merges the subarrays A [1 .. m] and A[m + 1 .. n],
assuming those subarrays are sorted in the input.

Proof: Let A[ 1 .. n] be any array and m any integer such that the subarrays
A [1 .. m ] and A[m +1 .. n] are sorted. We prove that for all k from 0 to n, the last
n — k — 1 iterations of the main loop correctly merge A[i .. m ] and A[j .. n] into

27

i. Recursion

B[k .. n]. The proof proceeds by induction on n — k + 1 , the number of elements
remaining to be merged.

If k > n, the algorithm correctly merges the two empty subarrays by doing
absolutely nothing. (This is the base case of the inductive proof.) Otherwise,
there are four cases to consider for the kth iteration of the main loop.

• If j > n, then subarray A[j .. n] is empty, so min (A[i.. m]uA[j..n])=A[i].

• If i > m, then subarray A[i.. m] is empty, so min (A[i.. m]uA[j..n])=A[j].

• Otherwise, if A[i] < A[j], then min (A[i.. m]uA[j..n])=A[i].

• Otherwise, we must have A[ i ] > A[j ], and min (A[ i.. m]uA[j..n])=A[j].

In all four cases, £>[/<] is correctly assigned the smallest element of A[i .. m] U
A[j .. n]. In the two cases with the assignment B[k] *— A[i], the Recursion Fairy
correctly merges—sorry, I mean the Induction Hypothesis implies that the last
n — k iterations of the main loop correctly merge A[i + 1 .. m] and A[j .. n] into
B[k + 1 .. n]. Similarly, in the other two cases, the Recursion Fairy also correctly
merges the rest of the subarrays. □

Theorem 1.2. MergeSort correctly sorts any input arrayA[ 1 .. n].

Proof: We prove the theorem by induction on n. If n < 1 , the algorithm
correctly does nothing. Otherwise, the Recursion Fairy correctly sorts—sorry, I
mean the induction hypothesis implies that our algorithm correctly sorts the
two smaller subarrays A[ 1 .. m] and A[m + 1 .. n ], after which they are correctly
MERGEd into a single sorted array (by Lemma 1.1). □

Analysis

Because the MergeSort algorithm is recursive, its running time is naturally
expressed as a recurrence. Merge clearly takes O(n) time, because it’s a simple
for-loop with constant work per iteration. We immediately obtain the following
recurrence for MergeSort:

T(n) = T(jn/21) + r(Ln/2j) + O(n).

As in most divide-and-conquer recurrences, we can safely strip out the floors
and ceilings (using a technique called domain transformations described later
in this chapter), giving us the simpler recurrence T(n) = 2 T(n/ 2 ) + O(n). The
“all levels equal” case of the recursion tree method (also described later in this
chapter) immediately implies the closed-form solution T(n) = O(nlogn). Even
if you are not (yet) familiar with recursion trees, you can verify the solution
T(n) = 0 (n log n) by induction.

28

1.5. Quicksort

1.5 Quicksort

Quicksort is another recursive sorting algorithm, discovered by Tony Hoare in
1959 and first published in 1961. In this algorithm, the hard work is splitting the
array into subsets so that merging the final result is trivial.

1. Choose a pivot element from the array.

2. Partition the array into three subarrays containing the elements smaller
than the pivot, the pivot element itself, and the elements larger than the
pivot.

3. Recursively quicksort the first and last subarrays.

Input:

S

0

R

T

I

N

G

E

X

A

M

p

L

Choose a pivot:

S

0

R

T

I

N

G

E

X

A

M

0

L

Partition:

A

G

0

E

I

N

L

M

0

T

X

s

R

Recurse Left:

A

E

G

I

L

M

N

0

0

T

X

s

R

Recurse Right:

A

E

G

I

L

M

N

0

0

R

s

T

X

Figure 1.7. A quicksort example.

A more detailed description of the algorithm is given in Figure 1.8. In the
separate Partition subroutine, the input parameter p is index of the pivot
element in the unsorted array; the subroutine partitions the array and returns
the new index of the pivot. There are many different efficient partitioning
algorithms; the one I’m presenting here is attributed to Nico Lomuto. 6 The
variable l counts the number of items in the array that are less than the pivot
element.

Figure 1.8. Quicksort

OuickSortIAI 1.. nl):

if (n > 1)

Choose a pivot element A[p]

r <— Partition(A,p)

QuickSort(A[ 1 .. r — 1])

({Recurse!))

QuiCKSoRT(A[r + 1.. n])

((Recurse!))

Partition(A[ 1.. n ], p ):

swap A[p]

i <— 0 ((#items < pivot))

for i <— 1 to n — 1
if A[i] < A[n]

£ <-£ + 1
swap A[£]«—»A[i]
swapA[l] ♦—> A[£ + 1 ]
return l + l

Correctness

Just like mergesort, proving that Quicksort is correct requires two separate
induction proofs: one to prove that Partition correctly partitions the array, and

6 Hoare proposed a more complicated “two-way” partitioning algorithm that has some
practical advantages over Lomuto’s algorithm. On the other hand, Hoare’s partitioning algorithm
is one of the places off-by-one errors go to die.

29

i. Recursion

the other to prove that Quicksort correctly sorts assuming Partition is correct.
To prove Partition is correct, we need to prove the following loop invariant: At
the end of each iteration of the main loop, everything in the subarray A [1 ..£]
is £ess than A[n], and nothing in the subarray A[£ + 1 .. i] is less than A[n],
I’ll leave the remaining straightforward but tedious details as exercises for the
reader.

Analysis

The analysis of quicksort is also similar to that of mergesort. Partition clearly
runs in O(n) time, because it’s a simple for-loop with constant work per iteration.
For Quicksort, we get a recurrence that depends on r, the rank of the chosen
pivot element:

T(n) = T(r - 1 ) + T(n - r) + O(n)

If we could somehow always choose the pivot to be the median element of the
array A, we would have r = \n/ 2 ], the two subproblems would be as close to
the same size as possible, the recurrence would become

T(n) = 2 T(jn/ 2 l-l) + r(|_n/ 2 j) + 0 (n) < 2 T(n/ 2 ) +O(n),

and we’d have T(n) = 0 (n log n) by the recursion tree method.

In fact, as we will see later in this chapter, we can actually locate the
median element in an unsorted array in linear time, but the algorithm is fairly
complicated, and the hidden constant in the O(-) notation is large enough to
make the resulting sorting algorithm impractical. In practice, most programmers
settle for something simple, like choosing the first or last element of the array.
In this case, r can take any value between 1 and n, so we have

T(n) = max (T(r — 1 ) + T[n — r) + O(n)).

l<r<rz v J

In the worst case, the two subproblems are completely unbalanced—either r = 1
or r = n —and the recurrence becomes T(n) < T{n — 1 ) + O(n). The solution is
T(n) = 0(n 2 ).

Another common heuristic is called “median of three”—choose three ele¬
ments (usually at the beginning, middle, and end of the array), and take the
median of those three elements the pivot. Although this heuristic is somewhat
more efficient in practice than just choosing one element, especially when
the array is already (nearly) sorted, we can still have r = 2 or r = n — 1 in
the worst case. With the median-of-three heuristic, the recurrence becomes
T(n) < T(l) + T(ji — 2 ) + O(n), whose solution is still T{n) = 0 (n 2 ).

Intuitively, the pivot element should "usually" fall somewhere in the middle
of the array, say between n /10 and 9 n/ 10 . This observation suggests that

30

1.6. The Pattern

the average-case running time should be 0 (n log n). Although this intuition is
actually correct, it’s correct only under the completely unrealistic assumption
that all permutations of the input array are equally likely. Real world data may
be random, but it is not random in any way that you can predict in advance,
and it is certainly not uniform! 7

Occasionally people consider “best case” running time for some reason. We
won’t.

i .6 The Pattern

Both mergesort and quicksort follow a general three-step pattern called divide
and conquer:

1. Divide the given instance of the problem into several independent smaller
instances.

2. Delegate each smaller instance to the Recursion Fairy.

3. Combine the solutions for the smaller instances into the final solution
for the given instance.

If the size of any instance falls below some constant threshold, we abandon
recursion and solve the problem directly, by brute force, in constant time.

Proving a divide-and-conquer algorithm correct almost always requires
induction. Analyzing the running time requires setting up and solving a
recurrence, which usually (but unfortunately not always!) can be solved using
recursion trees.

i .7 Recursion Trees

So what are these “recursion trees” I keep talking about? Recursion trees are
a simple, general, pictorial tool for solving divide-and-conquer recurrences. A
recursion tree is a rooted tree with one node for each recursive subproblem. The
value of each node is the amount of time spent on the corresponding subproblem
excluding recursive calls. Thus, the overall running time of the algorithm is the
sum of the values of all nodes in the tree.

To make this idea more concrete, imagine a divide-and-conquer algorithm
that spends 0(/(n)) time on non-recursive work, and then makes r recursive
calls, each on a problem of size n/c. Up to constant factors (which we can

7 On the other hand, if we choose the pivot index p uniformly at random, then Quicksort runs
in O(nlogn) time with high probability, for every possible input array. The key difference is that
the randomness is controlled by our algorithm, not by the all-powerful malicious adversary who
gives us input data after reading our code. The analysis of randomized quicksort is unfortunately
outside the scope of this book, but you can find relevant lecture notes at http://algorithms.wtf/.

31

i. Recursion

hide in the 0() notation), the running time of this algorithm is governed by the
recurrence

T(n) = r T{n/c) + f (n).

The root of the recursion tree for T(n) has value f{n) and r children,
each of which is the root of a (recursively defined) recursion tree for T(n/c).
Equivalently, a recursion tree is a complete r-ary tree where each node at depth d
contains the value f(n/c d ). (Feel free to assume that n is an integer power of c,
so that n/c d is always an integer, although in fact this doesn’t matter.)

In practice, I recommend drawing out the first two or three levels of the
tree, as in Figure 1.9.

+

r 1 - • f(n/c L )

Figure 1.9. A recursion tree for the recurrence T(n ) = r T(n/c) + /(n)

The leaves of the recursion tree correspond to the base case(s) of the
recurrence. Because we’re only looking for asymptotic bounds, the precise base
case doesn’t actually matter; we can safely assume T(n ) = 1 for all n < n 0 ,
where n 0 is an arbitrary positive constant. In particular, we can choose whatever
value of n 0 is most convenient for our analysis. For this example, I’ll choose
n o = 1 -

Now T(n) is the sum of all values in the recursion tree; we can evaluate this
sum by considering the tree level-by-level. For each integer i, the ith level of
the tree has exactly r l nodes, each with value /(n/c 1 ). Thus,

L

T(n) = J]r l f(n/c i )
i=0

(£)

where L is the depth of the tree. Our base case n 0 = 1 immediately implies
L = log c n, because n/c L = n () = 1 . It follows that the number of leaves in
the recursion tree is exactly r L = r logf n = n logc r . Thus, the last term in the
level-by-level sum (E) is n logcr •/(!) = 0(n logc ' ), because /(1) = 0(1).

32

1-7. Recursion Trees

There are three common cases where the level-by-level sum (E) is especially
easy to evaluate:

• Decreasing: If the sum is a decreasing geometric series —every term is a
constant factor smaller than the previous term—then T(n) = 0(/(n)). In
this case, the sum is dominated by the value at the root of the recursion tree.

• Equal: If all terms in the sum are equal, we immediately have T(n) =
0(f(n)-L) = 0(/(n)logn). (The constant c vanishes into the 0() notation.)

• Increasing: If the sum is an increasing geometric series —every term is a
constant factor larger than the previous term—then T(n) = O(n logcr ). In
this case, the sum is dominated by the number of leaves in the recursion
tree.

In the first and third cases, only the largest term in the geometric series matters;
all other terms are swallowed up by the O(-) notation. In the decreasing case,
we don’t even have to compute L; the asymptotic upper bound would still hold
if the recursion tree were infinite!

For example, if we draw out the first few levels of the recursion tree for the
(simplified) mergesort recurrence T(n) = 2T(n/2) + O(n), we discover that all
levels are equal, which immediately implies T(n) = O(nlogn).

The recursion tree technique can also be used for algorithms where the
recursive subproblems have different sizes. As an extreme example, the worst-
case recurrence for quicksort T(n) = T(n—l)+T(l)+0(n) gives us a completely
unbalanced recursion tree, where one child of each internal node is a leaf. The
level-by-level sum doesn’t fall into any of our three default categories, but we
can still derive the solution T{n) = 0(n 2 ) by observing that every level value is
at most n and there are at most n levels. (Moreover, since n/2 levels each have
value at least n/2, this analysis can be improved by at most a constant factor,
which for our purposes means not at all.)

Figure 1.10. The recursion trees for mergesort and quicksort

33

i. Recursion

^Ignoring Floors and Ceilings Is Okay, Honest

Careful readers might object that our analysis brushes an important detail under
the rug. The running time of mergesort doesn’t really obey the recurrence
T(n) = 2T(n/2) + 0(n); after all, the input size n might be odd, and what could
it possibly mean to sort an array of size 42^ or 17|? The actual mergesort
recurrence is somewhat messier:

T(n) = T{jn/2]) + T([n/2\) + O(n).

Sure, we could check that T(n) = 0(n log n) using induction, but the necessary
calculations would be awful. Fortunately, there is a simple technique for
removing floors and ceilings, called a domain transformation.

• First, we overestimate T(n), once by pretending that the two subproblem
sizes are equal, and again to eliminate the ceiling: 8

r(n) < 2T(\n/2]) + n < 2T(n/2 + l) + n.

• Second, we define a new function S(n) = T(n+a), choosing the constant a so
that S(n) satisfies the simpler recurrence of the form S(n) < 2S(n/2) + 0(n).
To find the correct constant a, we derive a recurrence for S from our given
recurrence for T:

S(n ) = T(n + a) [definition of S]

< 2T{n/2 + a/2 + 1) + n + a [recurrence for T]

= 2S{n/2 — a/2 + 1) + n + a [definition of S]

Setting a = 2 simplifies this recurrence to S(n) < 2S(n/2) + n + 2, which is
exactly what we wanted.

• Finally, the recursion tree method implies S(n) = 0(n log n), and therefore

T(n) = S(n —2) = 0((n — 2)log(n — 2)) = O(nlogn),
exactly as promised.

Similar domain transformations can be used to remove floors, ceilings, and even
lower order terms from any divide and conquer recurrence. But now that we
realize this, we don’t need to bother grinding through the details ever again!
From now on, faced with any divide-and-conquer recurrence, I will silently
brush floors and ceilings and lower-order terms under the rug, and I encourage
you to do the same.

8 Formally, we are treating T as a function over the reals, not just over the integers, that
satisfies the given recurrence with the base case T(n) = C for all n < n 0 , for some real numbers
C > 0 and rc 0 > 0 whose values don’t matter. If n happens to be an integer, then T(n) coincides
with the running time of an algorithm on an input of size n, but that doesn’t matter, either.

34

* 1 . 8 . Linear-Time Selection

*1.8 Linear-Time Selection

During our discussion of quicksort, I claimed in passing that we can find the
median of an unsorted array in linear time. The first such algorithm was
discovered by Manuel Blum, Bob Floyd, Vaughan Pratt, Ron Rivest, and Bob
Tarjan in the early 1970 s. Their algorithm actually solves the more general
problem of selecting the kth largest element in an n-element array, given the
array and the integer k as input, using a variant of an algorithm called quickselect
or one-armed quicksort. Quickselect was first described by Tony Hoare in 1961 ,
literally on the same page where he first published quicksort.

The generic quickselect algorithm chooses a pivot element, partitions the
array using the same Partition subroutine as Quicksort, and then recursively
searches only one of the two subarrays. Crucially, the correctness of this algorithm
does not depend on how the pivot is chosen.

QuickSelect(A[ 1.. n], k ):

if n = 1

return A[i]

else

Choose a pivot element A[p]
r <— Partition(A[1 .. n],p)

if k < r

return QuickSelect(A[1 .. r — 1], k)
else if k > r

return QuickSelect(A[> + 1 .. n], k — r)

else

return A[r]

The worst-case running time of Hoare’s QuickSelect algorithm obeys a
recurrence similar to Quicksort. We don’t know the value of r or which of the
two subarrays we’ll recursively search, so we’ll just assume the worst.

T(n) < max max{T(r — 1), T(n — r)} + O(n)

l<r<n

We can simplify the recurrence slightly by letting i denote the length of the
recursive subproblem:

T(n) < max T(£) + 0(n)

0<t<n—1

If the chosen pivot element is always either the smallest or largest element in
the array, the recurrence simplifies to T(n) = T{n — 1) + O(n), which implies
T(n) = 0(n 2 ). (The recursion tree for this recurrence is just a simple path.)

We could avoid this quadratic worst-case behavior if we could somehow
magically choose a good pivot, meaning £ < an for some constant a < 1. In

35

i. Recursion

this case, the recurrence would simplify to

T(n) < T(an) + O(n).

This recurrence expands into a descending geometric series, which is dominated
by its largest term, so T(n) = O(n). (Again, the recursion tree is just a simple
path. The constant in the O(n) running time depends on the constant a.)

In other words, if we could somehow quickly find an element that’s even
close to the median in linear time, we could find the exact median in linear
time. So now all we need is an Approximate Median Fairy. The Blum-Floyd-
Pratt-Rivest-Tarjan algorithm chooses a good quickselect pivot by recursively
computing the median of a carefully-chosen subset of the input array. The
Approximate Median Fairy is just the Recursion Fairy in disguise!

Specifically, we divide the input array into \n/ 5] blocks, each containing
exactly 5 elements, except possibly the last. (If the last block isn’t full, just throw
in a few oos.) We compute the median of each block by brute force, collect
those medians into a new array M[1.. [n/5]], and then recursively compute the
median of this new array. Finally we use the median of medians (called “mom”
in the following pseudocode) as the pivot in one-armed quicksort.

MomSelect(A[ 1.. nl, k):

if n < 25 ((or whatever))

use brute force

else

m <— fn/5]

for i «— 1 to m

M[i]<— MedianOfFive(A[5i — 4 .. 5i])

((Brute force!))

mom <— MomSelect(M[1 .. m], Lm/2j)

((Recursion!))

r <— Partition(A[ 1 ..n],mom)

if k < r

return MomSelect(A[1 .. r — 1], k)

((Recursion!))

else if k > r

return MomSelect(A[> + 1.. n], k — r)

((Recursion!))

else

return mom

MomSelect uses recursion for two different purposes; once to choose a
pivot element (mom), and the second time to search through the entries on one
side of that pivot.

36

Analysis

But why is this fast? The first key insight is that the median of medians is a
good pivot. Mom is larger than |_[n/5]/2j — lf«n/10 block medians, and each

* 1 . 8 . Linear-Time Selection

block median is larger than two other elements in its block. Thus, mom is bigger
than at least 3n/10 elements in the input array; symmetrically, mom is smaller
than at least 3n/10 elements. Thus, in the worst case, the second recursive call
searches an array of size at most 7n/ 10.

We can visualize the algorithm’s behavior by drawing the input array as a
5 x [n/5] grid, which each column represents five consecutive elements. For
purposes of illustration, imagine that we sort every column from top down, and
then we sort the columns by their middle element. (Let me emphasize that the
algorithm does not actually do this!) In this arrangement, the median-of-medians
is the element closest to the center of the grid.

□□□□□□□□□□□□□□□□
OOO0OOO00OO00CXI]
GOOGOGOGOOGGOS

pppppppppppppppp

The left half of the first three rows of the grid contains 3n/10 elements, each
of which is smaller than mom. If the element we’re looking for is larger than
mom, our algorithm will throw away everything smaller than mom, including
those 3n/10 elements, before recursing. Thus, the input to the recursive
subproblem contains at most 7n /10 elements. A symmetric argument implies
that if our target element is smaller than mom, we discard at least 3n/10
elements larger than mom, so the input to our recursive subproblem has at most
7n/10 elements.

OOOOOOOOOOOOOOOO

oooooooco □□□□□□□
□ o ooo c oofirnoorno
0000000000000000
oooooooooooooooo

Okay, so mom is a good pivot, but our algorithm still makes two recursive
calls instead of just one; how do we prove linear time? The second key insight is
that the total size of the two recursive subproblems is a constant factor smaller
than the size of the original input array. The worst-case running time of the
algorithm obeys the recurrence

T(n) < T(n/5) + T(7n/10) + 0(n).

If we draw out the recursion tree for this recurrence, we observe that the total
work at each level of the recursion tree is at most 9/10 the total work at the

37

i. Recursion

previous level. Thus, the level-by-level sum is a descending geometric series,
giving us the solution T(n) = O(n). Hooray! Thanks, Mom!

Figure i .11 . The recursion trees for MomSelect and a similar selection algorithm with blocks of size 3

Sanity Checking

At this point, many students ask about that magic constant 5. Why did we
choose that particular block size? The answer is that 5 is the smallest odd block
size that gives us a descending geometric series in the running time! (Even block
sizes introduce additional complications.) If we had used blocks of 3 elements
instead, the running-time recurrence becomes

T(n) < T(n/3) + T(2n/3) + O(n).

In this case, every level of the recursion tree has the same value n. The leaves
of the recursion tree are not all at the same level, but for purposes of deriving
an upper bound, it suffices to observe that the deepest leaf is at level log 3 / 2 n,
so the total work in the tree is at most 0(n log 3 / 2 n) = O(nlogn). On the other
hand, the shallowest leaf has depth log 3 n, so the analysis can’t be improved by
more than a constant factor. This algorithm is no faster than sorting!

Finer analysis reveals that the constant hidden by the 0() notation is quite
large, even if we count only comparisons. Selecting the median of 5 elements
requires at most 6 comparisons, so we need at most 6n/5 comparisons to set
up the recursive subproblem. Naively partitioning the array after the recursive
call would require n — 1 comparisons, but we already know 3n/10 elements
larger than the pivot and 3n/10 elements smaller than the pivot, so partitioning
actually requires only 2n/5 additional comparisons. Thus, a more precise
recurrence for the worst-case number of comparisons is

T(n) < T(n/5) + T(7n/ 10) + 8n/5.

The recursion tree method implies the upper bound

T{n)<

8 n

yf-

5 1^ 10

8 n

10 = 16 n.

38

i.g. Fast Multiplication

In practice, this algorithm isn’t as horrible as this worst-case analysis pre¬
dicts—getting a worst-case partition at every level of recursion is incredibly
unlikely—but it is slower than sorting for even moderately large arrays. 9

i.g Fast Multiplication

In the previous chapter, we saw two different algorithms for multiplying two
n-digit numbers in 0(n 2 ) time: the grade-school lattice algorithm and the
Egyptian peasant algorithm.

Maybe we can get a more efficient algorithm by splitting the digit arrays in
half and exploiting the following identity:

(10 m a + b)(10 m c + d) = 10 2m ac + 10 m (bc + ad) + bd

This recurrence immediately suggests the following divide-and-conquer algo¬
rithm to multiply two n-digit numbers x and y. Each of the four sub-products
ac, be, ad, and bd is computed recursively, but the multiplications in the last
line are not recursive, because we can multiply by a power of ten by shifting the
digits to the left and filling in the correct number of zeros, all in O(n) time.

Multiply(x, y, n):

if n = 1

return x ■ y

else

m <— [n/2j

a <— |x/10 m J; b <— x mod 10 m

((x = 10 m a + b))

c <— Ly/10 m J; d y mod 10 m
e <— MuLTiPLY(a, c, m)
f <— MuLTiPLY(b, d, m)
g MULTIPLY(b, c, m)

h <— MuLTiPLY(a, d, m)
return 10 2m e + 10 m (g +h) + f

{(y = 10 m c + d))

Correctness of this algorithm follows easily by induction. The running time for
this algorithm follows the recurrence

r(fi) = 4r(rn/2l) + 0(n).

The recursion tree method transforms this recurrence into an increasing geo¬
metric series, which implies T(n) = 0(n log24 ) = 0(n 2 ). In fact, this algorithm
multiplies each digit of x with each digit of y, just like the lattice algorithm.
So I guess that didn’t work. Too bad. It was a nice idea.

9 In fact, the right way to choose the pivot element in practice is to choose it uniformly at
random. Then the expected number of comparisons required to find the median is at most 4 n.
See my randomized algorithms lecture notes http://algorithms.wtf for more details.

39

i. Recursion

z7T\ /7v\ /A\ /7v\

77/4

71/4

71/4

77/4

77/4

77/4

77/4

77/4

77/4

77/4

77/4

77/4

77/4

77/4

77/4

77/4

Figure 1.12. The recursion tree for naive divide-and-conquer multiplication

In the mid- 1950 s, Andrei Kolmogorov, one of the giants of 20 th century
mathematics, publicly conjectured that there is no algorithm to multiply two
n-digit numbers in o(n 2 ) time. Kolmogorov organized a seminar at Moscow Uni¬
versity in i 960 , where he restated his “n 2 conjecture” and posed several related
problems that he planned to discuss at future meetings. Almost exactly a week
later, a 23 -year-old student named Anatoli! Karatsuba presented Kolmogorov
with a remarkable counterexample. According to Karastuba himself,

After the seminar I told Kolmogorov about the new algorithm and about the
disproof of the n 2 conjecture. Kolmogorov was very agitated because this
contradicted his very plausible conjecture. At the next meeting of the semi¬
nar, Kolmogorov himself told the participants about my method, and at that
point the seminar was terminated.

Karastuba observed that the middle coefficient bc+ad can be computed from the
other two coefficients ac and bd using only one more recursive multiplication,
via the following algebraic identity:

ac + bd — (a — b)(c — d) = be + ad

This trick lets us replace the four recursive calls in the previous algorithm with
just three recursive calls, as shown below:

FastMultiply(x, y, n):

if 71 = 1

return x ■ y

else

m <— \n/2\

a <— [x/10 m J; b <— x mod 10 m ((x = 10 m a + b})

c <— Ly/10 m J; d <— y mod 10 m ({y = 10 m c + d))

e <— FASTMuLTiPLY(a, c, m)

/ <— FASTMULTIPLY(b, d. Til)
g <— FastMultiplyCo — b,c — d,m)
return 10 2m e + 10 m (e+/-g)+/

40

The running time of Karatsuba’s FastMultiply algorithm follows the recurrence

T(n) < 3T([n/2]) + 0(n)

i.g. Fast Multiplication

Again, we can safely ignore the ceiling in the recursive argument, and the
recursion tree method transforms the recurrence into an increasing geometric
series, but the new solution is only T(n) = 0(n log23 ) = 0(n 1-58496 ), a significant
improvement over our earlier quadratic-time algorithm . 10 Karatsuba’s algorithm
arguably launched the design and analysis of algorithms as a formal field of
study.

Figure 1.13. The recursion tree for Karatsuba's divide-and-conquer multiplication algorithm

We can take Karatsuba’s idea even further, splitting the numbers into more
pieces and combining them in more complicated ways, to obtain even faster
multiplication algorithms. Andrei Toom and Stephen Cook discovered an infinite
family of algorithms that split any integer into k parts, each with n/k digits, and
then compute the product using only 2k — 1 recursive multiplications. For any
fixed k, the resulting algorithm runs in 0 (n 1+1 ^ lg,c - ) ) time, where the hidden
constant in the O(-) notation depends on k.

Ultimately, this divide-and-conquer strategy led Gauss (yes, really) to the
discovery of the Fast Fourier transform , u The basic FFT algorithm itself runs
in O(nlogn) time; using FFTs for integer multiplication incurs some small
additional overhead. The fastest integer multiplication algorithm currently
known, published by David Harvey and Joris van der Hoeven in 2018 , runs in
0{n log n 4 log n ) time. Here, log* n denotes the slowly growing iterated logarithm
of n, which is the number of times one must take the logarithm of n before the
value is less than 1 :

* f 1 if n < 2,

lg* n = f

[ 1 + lg*(lg n) otherwise.

For all practical purposes, log* n < 6. It has been conjectured that the best
possible algorithm for multiply two n-digit numbers runs in 0(nlogn) time.

10 My presentation simplifies the actual history slightly. In fact, Karatsuba proposed an
algorithm based on the formula (a + c)(b + d) — ac — bd = be + ad. This algorithm also runs
in 0(n lg3 ) time, but the actual recurrence is slightly messier: a — b and c — d are still m-digit
numbers, but a + b and c + d might each have m + 1 digits. The simplification presented here is
due to Donald Knuth.

u See http://algorithms.wtf for lecture notes on Fast Fourier transforms.

41

i. Recursion

1.10 Exponentiation

Given a number a and a positive integer n, suppose we want to compute a n . The
standard naive method is a simple for-loop that performs n — 1 multiplications
by a:

SlqwPower(q, n):
x <— a

for i «— 2 to n
x <— x • a
return x

This iterative algorithm requires n multiplications.

The input parameter a could be an integer, or a rational, or a floating point
number. In fact, it doesn’t need to be a number at all, as long as it’s something
that we know how to multiply. For example, the same algorithm can be used
to compute powers modulo some finite number (an operation commonly used
in cryptography algorithms) or to compute powers of matrices (an operation
used to evaluate recurrences and to compute shortest paths in graphs). Because
we don’t know what kind of object we’re multiplying, we can’t know how much
time a single multiplication requires, so we’re forced to analyze the running
time in terms of the number of multiplications.

There is a much faster divide-and-conquer method, originally proposed by
the Indian prosodist Pingala in the 2 nd century bce, which uses the following
simple recursive formula:

r

1

a

n _

< ( a "/ 2 ) 2
(a Ln/2j ) 2

a

if n = 0

if n > 0 and n is even
otherwise

PingalaPower(ci, n):

if n

= 1

return a

else

x <— PingalaPower(<i, \_n/2\)

if n is even

return x • x

else

return x ■ x ■ a

The total number of multiplications performed by this algorithm satisfies the
recurrence T(n) < T(n/2) + 2. The recursion-tree method immediately give us
the solution T(n) = O(logn).

A nearly identical exponentiation algorithm can also be derived directly
from the Egyptian peasant multiplication algorithm from the previous chapter,

42

Exercises

by replacing addition with multiplication (and in particular, replacing duplation
with squaring).

( 1 if n = 0

(fl ^) n /2 if n > 0 and n is even
(a 2 )I"/ 2 ! • a otherwise

PeasantPowerCq, n):
if ti= 1

return a
else if n is even

return PEASANTPowER(a 2 ,n/2)

else

return PeasantPower(h 2 , [n/2j) • a

This algorithm—which might reasonably be called “squaring and mediation”—
also performs only 0(log n) multiplications. Both of these algorithms are
asymptotically optimal—any algorithm that computes a n must perform at least
fl(log n) multiplications.

In fact, when n is a power of two, both of these algorithm are exactly optimal.
However, there are slightly faster methods for other values of n. For example,
PingalaPower and PeasantPower each compute a 15 using six multiplications,
but in fact only five multiplications are necessary:

• Pingala: a —> a 2 —> a 3 —> a 6 —> a 7 —> a 14 —» a 15

• Peasant: a —» a 2 —» a 4 —» a 8 —» a 12 —> a 14 —» a 15

• Optimal: a —> a 2 —> a 3 —> a 5 —> a 10 —> a 15

It is a long-standing open question whether the absolute minimum number of
multiplications for a given exponent n can be computed efficiently.

Exercises

Tower of Hanoi

l. Prove that the original recursive Tower of Hanoi algorithm performs exactly
the same sequence of moves—the same disks, to and from the same pegs,
in the same order—as each of the following non-recursive algorithms. The
pegs are labeled 0 , 1 , and 2 , and our problem is to move a stack of n disks
from peg 0 to peg 2 (as shown on page 24 ).

(a) If n is even, swap pegs 1 and 2. At the ith step, make the only legal
move that avoids peg i mod 3. If there is no legal move, then all disks
are on peg i mod 3, and the puzzle is solved.

43

i. Recursion

(b) For the first move, move disk 1 to peg 1 if n is even and to peg 2 if n is
odd. Then repeatedly make the only legal move that does not undo the
previous move. If no such move exists, the puzzle is solved.

(c) Pretend that disks n + 1, n + 2, and n + 3 are at the bottom of pegs 0, 1,
and 2, respectively. Repeatedly make the only legal move that satisfies
the following constraints, until no such move is possible.

• Do not place an odd disk directly on top of another odd disk.

• Do not place an even disk directly on top of another even disk.

• Do not undo the previous move.

(d) Let p(n) denote the smallest integer k such that n/2 k is not an integer.
For example, p(42) = 2, because 42/2 1 is an integer but 42/2 2 is not.
(Equivalently, p(n) is one more than the position of the least significant
1 in the binary representation of n .) Because its behavior resembles the
marks on a ruler, p(n) is sometimes called the ruler function.

HANOi(n):

i <- 1

while p(i) < n

if n — i is even

move disk p ( i ) forward

«0-»l-»2-»0))

else

move disk p(i) backward

((0 —> 2 —> 1 —> 0))

i <—i + 1

2 . The Tower of Hanoi is a relatively recent descendant of a much older
mechanical puzzle known as the Chinese linked rings, Baguenaudier, Car¬
dan’s Rings, Meleda, Patience, Tiring Irons, Prisoner’s Lock, Spin-Out, and
many other names. This puzzle was already well known in both China
and Europe by the 16 th century. The Italian mathematician Luca Pacioli
described the 7 -ring puzzle and its solution in his unpublished treatise De
Viribus Quantitatis, written between 1498 and 1506; 12 only a few years later,
the Ming-dynasty poet Yang Shen described the 9 -ring puzzle as “a toy for
women and children”. The puzzle is apocryphally attributed to a 2 nd-century
Chinese general, who gave the puzzle to his wife to occupy her time while
he was away at war.

The Baguenaudier puzzle has many physical forms, but one of the most
common consists of a long metal loop and several rings, which are connected
to a solid base by movable rods. The loop is initially threaded through the

12 De Viribus Quantitatis [On the Powers of Numbers] is an important early work on recreational
mathematics and perhaps the oldest surviving treatise on magic. Pacioli is better known for
Summa de Aritmetica, a near-complete encyclopedia of late 15 th-century mathematics, which
included the first description of double-entry bookkeeping.

44

Exercises

76543a 1

Figure 1.14. The 7-ring Baguenaudier, from Recreations Mathematiques by Edouard Lucas (1891) (See
Image Credits at the end of the book.)

rings as shown in the figure above; the goal of the puzzle is to remove the
loop.

More abstractly, we can model the puzzle as a sequence of bits, one for
each ring, where the ith bit is 1 if the loop passes through the ith ring and
0 otherwise. (Here we index the rings from right to left, as shown in the
figure.) The puzzle allows two legal moves:

• You can always flip the 1 st C— rightmost) bit.

• If the bit string ends with exactly z 0s, you can flip the (z + 2)th bit.

The goal of the puzzle is to transform a string of n 1 s into a string of n 0s.
For example, the following sequence of 21 moves solves the 5 -ring puzzle:

11111 4 11 H 0 4 11010 4 non 4 11001 4 11000

4 01000 4 01001 4 01011 4 01010 4 01110

4 011H 401101 4 0ii00 4 00i00 400i0i

4 00111 4 00110 4 00010 4 00011 4 00001 4 00000

* (a) Call a sequence of moves reduced if no move is the inverse of the previous
move. Prove that for any non-negative integer n, there is exactly one
reduced sequence of moves that solves the n-ring Baguenaudier puzzle.
[Hint: This problem is much easier if you’re already familiar with
graphs.]

(b) Describe an algorithm to solve the Baguenaudier puzzle. Your input is
the number of rings n; your algorithm should print a reduced sequence
of moves that solves the puzzle. For example, given the integer 5 as
input, your algorithm should print the sequence 1 ,3, 1 , 2 , 1 ,5, 1 , 2 , 1 ,3,
1,2,1,4,1,2,1,3,1,2,1.

(c) Exactly how many moves does your algorithm perform, as a function of
n? Prove your answer is correct.

45

i. Recursion

3 . A less familiar chapter in the Tower of Hanoi’s history is its brief relocation
of the temple from Benares to Pisa in the early 13 th century. The relocation
was organized by the wealthy merchant-mathematician Leonardo Fibonacci,
at the request of the Holy Roman Emperor Frederick II, who had heard
reports of the temple from soldiers returning from the Crusades. The Towers
of Pisa and their attendant monks became famous, helping to establish Pisa
as a dominant trading center on the Italian peninsula.

Unfortunately, almost as soon as the temple was moved, one of the
diamond needles began to lean to one side. To avoid the possibility of
the leaning tower falling over from too much use, Fibonacci convinced the
priests to adopt a more relaxed rule: Any number of disks on the leaning
needle can be moved together to another needle in a single move. It was
still forbidden to place a larger disk on top of a smaller disk, and disks had to
be moved one at a time onto the leaning needle or between the two vertical
needles.

^ /

J

1

J

^ j

l nb

J

L J

L ••• /

4

Figure 1.15. The Towers of Pisa. In the fifth move, two disks are taken off the leaning needle.

Thanks to Fibonacci’s new rule, the priests could bring about the end of
the universe somewhat faster from Pisa then they could than could from
Benares. Fortunately, the temple was moved from Pisa back to Benares after
the newly crowned Pope Gregory IX excommunicated Frederick II, making
the local priests less sympathetic to hosting foreign heretics with strange
mathematical habits. Soon afterward, a bell tower was erected on the spot
where the temple once stood; it too began to lean almost immediately . 13

Describe an algorithm to transfer a stack of n disks from one vertical
needle to the other vertical needle, using the smallest possible number of
moves. Exactly how many moves does your algorithm perform?

4 . Consider the following restricted variants of the Tower of Hanoi puzzle. In
each problem, the pegs are numbered 0 , 1 , and 2 , and your task is to move
a stack of n disks from peg 0 to peg 2 , exactly as in problem 1 .

(a) Suppose you are forbidden to move any disk directly between peg 1 and
peg 2; every move must involve peg 0. Describe an algorithm to solve

13 Portions of this story are actually true.

46

Exercises

this version of the puzzle in as few moves as possible. Exactly how many
moves does your algorithm make?

* r (b) Suppose you are only allowed to move disks from peg 0 to peg 2, from
peg 2 to peg 1, or from peg 1 to peg 0. Equivalently, suppose the pegs
are arranged in a circle and numbered in clockwise order, and you are
only allowed to move disks counterclockwise. Describe an algorithm to
solve this version of the puzzle in as few moves as possible. How many

fH

Q

Q

Q

© 0

* -K§)

( 0 (§)

5

6

7

8 ^^

Figure i .16. The first several moves in a counterclockwise Towers of Hanoi solution.

**(c) Finally, suppose your only restriction is that you may never move a disk
directly from peg 0 to peg 2. Describe an algorithm to solve this version
of the puzzle in as few moves as possible. How many moves does your
algorithm make? [Hint: Matrices! This variant is considerably harder
to analyze than the other two.]

5 . Consider the following more complex variant of the Tower of Hanoi puzzle.
The puzzle has a row of k pegs, numbered from 1 to k. In a single turn, you
are allowed to move the smallest disk on peg i to either peg i — 1 or peg i +1 ,
for any index i; as usual, you are not allowed to place a bigger disk on a
smaller disk. Your mission is to move a stack of n disks from peg 1 to peg k.

(a) Describe a recursive algorithm for the case k = 3. Exactly how many
moves does your algorithm make? (This is exactly the same as problem
4(a).)

(b) Describe a recursive algorithm for the case k = n + 1 that requires at
most 0(n 3 ) moves. [Hint: Use part (a).]

r (c) Describe a recursive algorithm for the case k = n + 1 that requires at
most 0(n 2 ) moves. [Hint: Don’t use part (a).]

v (d) Describe a recursive algorithm for the case k = Jn that requires at most
a polynomial number of moves. (What polynomial??)

47

i. Recursion

*(e) Describe and analyze a recursive algorithm for arbitrary n and k. How
small must k be (as a function of n ) so that the number of moves is
bounded by a polynomial in n?

Recursion Trees

6 . Use recursion trees to solve each of the following recurrences.

A(n) = 2A(n/4) + Vn B{n) = 2B{n/A) + n C(n) = 2C(n/4) + n 2
D(n) = 3D(n/3) + Vn E{n) = 3E{n / 3) + n F(n) = 3F(n/3) + n 2
G(n) = 4G(n/2) + Vn H(n) = 4H(n/2) + n 7(n) = 4/(n/2) + n 2

7 . Use recursion trees to solve each of the following recurrences.

(j) J(n) = J(n/2) + J(n/3) + J(n/ 6 ) + n

(k) K(ri) = K{n/ 2) + 2JC(n/3) + 3JC(n/4) + n 2

(l) L(n) = L(n/15) + L(n/10) + 2L(n/6) +

v 8 . Use recursion trees to solve each of the following recurrences.

(m) M(n) = 2M(n/2) + O(nlogn)

(n) iV(n) = 2iV(n/2) + 0(n/logn)

(p) P{n) = i/nP{y/n) + n

(q) Q(n) = \/2nQ(\/2n)+ -/n

Sorting

9 . Suppose you are given a stack of n pancakes of different sizes. You want to
sort the pancakes so that smaller pancakes are on top of larger pancakes.
The only operation you can perform is a flip —insert a spatula under the
top k pancakes, for some integer k between 1 and n, and flip them all over.

48

Exercises

(a) Describe an algorithm to sort an arbitrary stack of n pancakes using
0(n) flips. Exactly how many flips does your algorithm perform in the
worst case ? 14 [Hint: This problem has nothing to do with the Tower of
Hanoi.]

(b) For every positive integer n, describe a stack of n pancakes that requires

flips to sort.

(c) Now suppose one side of each pancake is burned. Describe an algorithm
to sort an arbitrary stack of n pancakes, so that the burned side of every
pancake is facing down, using O(n) flips. Exactly how many flips does
your algorithm perform in the worst case?

to. Recall that the median of three heuristic examines the first, last, and middle
element of the array, and uses the median of those three elements as a
quicksort pivot. Prove that quicksort with the median-of-three heuristic
requires ff(n 2 ) time to sort an array of size n in the worst case. Specifically,
for any integer n, describe a permutation of the integers 1 through n,
such that in every recursive call to median-of-three-quicksort, the pivot is
always the second smallest element of the array. Designing this permutation
requires intimate knowledge of the Partition subroutine.

(a) As a warm-up exercise, assume that the Partition subroutine is stable,
meaning it preserves the existing order of all elements smaller than the
pivot, and it preserves the existing order of all elements smaller than
the pivot.

r (b) Assume that the Partition subroutine uses the specific algorithm listed
on page 29 , which is not stable.

11 . (a) Hey, Moe! Hey, Larry! Prove that the following algorithm actually sorts
its input!

STOOGESoRT(A[ 0 ..n— 11 ) :

if n = 2 and A[ 0 ] >A[ 1 ]

swap A[ 0 ]«—»A[l]

else if n > 2

m = [ 2 n/ 3 ]

StoogeSort(A [0 .. m —

1 ])

StoogeSort(A[h — m.

n—1])

StoogeSort(A [0 .. m —

1 ])

(b) Would StoogeSort still sort correctly if we replaced m = [2n/3] with
m = [2n/3j? Justify your answer.

14 The exact worst-case optimal number of flips required to sort n pancakes (either burned or
unburned) is an long-standing open problem; just do the best you can.

49

i. Recursion

(c) State a recurrence (including the base case(s)) for the number of
comparisons executed by StoogeSort.

(d) Solve the recurrence, and prove that your solution is correct. [Hint:
Ignore the ceiling.]

(e) Prove that the number of swaps executed by StoogeSort is at most Q).

12 . The following cruel and unusual sorting algorithm was proposed by Gary
Miller:

Cruel(A[ 1 .. nj):

if n> 1

Cruel(A [1 ..n/ 2 ])

Cruel(A[h /2 +1 ..n])

Unusual(A [1 ..n])

Unusual(A[~ 1 ..nl):

if n = 2

if A[l] > A[ 2] ({the only comparison!))

swap A[l]«—»A[2]

else

for i <— 1 to n/4 ((swap 2nd and 3rd quarters))

swap A[i + n/4] <—» A[i + n/2]

Unusual(A [1 .. n/ 2 ]) (( recurse on left half))

Unusual(A[r /2 + 1 .. n]) (( recurse on right half))

Unusual(A[r /4 + 1 .. 3 n/ 4 ]) ((recurse on middle half))

The comparisons performed by this algorithm do not depend at all on
the values in the input array; such a sorting algorithm is called oblivious.
Assume for this problem that the input size n is always a power of 2.

(a) Prove by induction that Cruel correctly sorts any input array. [Hint:
Consider an array that contains n/4 Is, n/4 2s, n/4 3s, and n/4 4s. Why
is this special case enough?]

(b) Prove that Cruel would not correctly sort if we removed the for-loop
from Unusual.

(c) Prove that Cruel would not correctly sort if we swapped the last two
lines of Unusual.

(d) What is the running time of Unusual? Justify your answer.

(e) What is the running time of Cruel? Justify your answer.

13 . An inversion in an array A[1.. n] is a pair of indices (i, j) such that i < j and
A[t] > A[j]. The number of inversions in an n-element array is between 0
(if the array is sorted) and Q) (if the array is sorted backward). Describe
and analyze an algorithm to count the number of inversions in an n-element
array in 0(n log n) time. [Hint: Modify mergesort.]

50

Exercises

14 . (a) Suppose you are given two sets of n points, one set {pi,P 2 > • • • ,p n } on the
line y = 0 and the other set {q\,q 2 , ■■■, q n } on the line y = 1. Create a set
of n line segments by connect each point p ; to the corresponding point q ; .
Describe and analyze a divide-and-conquer algorithm to determine how
many pairs of these line segments intersect, in 0(n log n) time. [Hint:
See the previous problem.]

(b) Now suppose you are given two sets {p\,P 2 , ■ ■ ■ ,P n ) and {<Ji, 0.2’ ■ ■ ■, <J n }
of n points on the unit circle. Connect each point p; to the corresponding
point q,. Describe and analyze a divide-and-conquer algorithm to
determine how many pairs of these line segments intersect in 0 (n log 2 n)
time. [Hint: Use your solution to part (a).]

r (c) Describe an algorithm for part (b) that runs in 0(n log n) time. [Hint:
Use your solution from part (b)!]

Figure 1.18. Eleven intersecting pairs of segments with endpoints on parallel lines, and ten intersecting
pairs of segments with endpoints on a circle.

15 . (a) Describe an algorithm that sorts an input array A[1..n] by calling a
subroutine SQRTSoRT(k), which sorts the subarray A\_k + 1.. k + -/n] in
place, given an arbitrary integer k between 0 and n — yfh as input. (To
simplify the problem, assume that ybi is an integer.) Your algorithm is
only allowed to inspect or modify the input array by calling SqrtSort;
in particular, your algorithm must not directly compare, move, or copy
array elements. How many times does your algorithm call SqrtSort in
the worst case?

*(b) Prove that your algorithm from part (a) is optimal up to constant factors.
In other words, if /(n) is the number of times your algorithm calls
SqrtSort, prove that no algorithm can sort using o(/(n)) calls to
SqrtSort.

(c) Now suppose SqrtSort is implemented recursively, by calling your
sorting algorithm from part (a). For example, at the second level of
recursion, the algorithm is sorting arrays roughly of size n 1 ' 4 . What
is the worst-case running time of the resulting sorting algorithm? (To

51

i. Recursion

simplify the analysis, assume that the array size n has the form 2 2 \ so
that repeated square roots are always integers.)

Selection

r 6 . Suppose we are given a set S of n items, each with a value and a weight. For
any element x € S, we define two subsets

• S <x is the set of elements of S whose value is less than the value of x.

• S >x is the set of elements of S whose value is more than the value of x.

For any subset Rt S, let w(R) denote the sum of the weights of elements in R.
The weighted median of R is any element x such that w(S <x ) < w(S)/2
and w(S >x ) < w(S)/2.

Describe and analyze an algorithm to compute the weighted median
of a given weighted set in O(n) time. Your input consists of two unsorted
arrays S[1.. n] and W[ 1.. n ], where for each index i, the ith element has
value S[i] and weight W[i], You may assume that all values are distinct and
all weights are positive.

17 . (a) Describe an algorithm to determine in O(n) time whether an arbitrary
array A[1.. n] contains more than n/4 copies of any value.

(b) Describe and analyze an algorithm to determine, given an arbitrary
array A[1.. n] and an integer k, whether A contains more than k copies
of any value. Express the running time of your algorithm as a function
of both n and k.

Do not use hashing, or radix sort, or any other method that depends
on the precise input values, as opposed to their order.

r 8 . Describe an algorithm to compute the median of an array A[ 1.. 5] of distinct
numbers using at most 6 comparisons. Instead of writing pseudocode,
describe your algorithm using a decision tree: A binary tree where each
internal node contains a comparison of the form “A[i] ^ A[j]?” and each
leaf contains an index into the array.

19. Consider the generalization of the Blum-Floyd-Pratt-Rivest-Tarjan Select
algorithm shown in Figure 1.20, which partitions the input array into \n/b]
blocks of size b, instead of [n/5] blocks of size 5, but is otherwise identical.

(a) State a recurrence for the running time of Mom & Select, assuming that
b is a constant (so the subroutine MedianOfB runs in 0(1) time). In
particular, how do the sizes of the recursive subproblems depend on the
constant b? Consider even b and odd b separately.

52

Exercises

(a[1]A[3])
< >

(a[1]:A[2])

< >

_zl ^

(a[1]A[3])
< >

(A[2]A[3]) ^ (A[2]A[3])

A[2] I A[3]

A[3] IA[2]

Figure 1.19. Finding the median of a 3-element array using at most 3 comparisons

MoMj,Select(A[1 .. n], k ):
if n<b 2

use brute force

else

m <— \n/b1
for i <— 1 to m

M[i] <— MEDiANOFB(A[b(i — 1) + 1.. bi])
mom b <— Mom^Select(M[1 .. m], Lm/2j)

r <— Partition(A[ 1 .. n], mom h )

if k < r

return Mom 5 Select(A[1 .. r — 1], fc)
else if k > r

return Mom^Select(A[)' + 1.. n], k — r)

else

return mom b

Figure 1.20. A parametrized family of selection algorithms; see problem ig.

(b) What is the worst-case running time of Mom 1 Select? [Hint: This is a
trick question.]

* r (c) What is the worst-case running time of Mom 2 Select? [Hint: This is an
unfair question!]

Y (d) What is the worst-case running time of Mom 3 Select? Finding an upper
bound on the running time is straightforward; the hard part is showing
that this analysis is actually tight. [Hint: See problem 10 .]

r (e) What is the worst-case running time of Mom 4 Select? Again, the hard
part is showing that the analysis cannot be improved. 15

15 The median of four elements is either the second smallest or the second largest. In 2014,
Ke Chen and Adrian Dumitrescu proved that if we modify Mom 4 Select to find second-smallest
elements when fc < n/2 and second-largest elements when k > n/ 2 , the resulting algorithm runs
in 0 (n ) time! See their paper “Select with Groups of 3 or 4 Takes Linear Time” (WADS 2015,
arXiv:i409.36oo) for details.

53

i. Recursion

(f) For any constants b > 5, the algorithm Mom ; , Select runs in O(n) time,
but different values of b lead to different constant factors. Let M{b)
denote the minimum number of comparisons required to find the median
of b numbers. The exact value of M(b) is known only for b < 13:

b

1 2 3

4

5

6

7

8

9

10

11

12

13

M(b)

0 13

4

6

8

10

12

14

16

18

20

23

For each b between 5 and 13, find an upper bound on the running time
of Mom;,Select of the form T(n) < a b n for some explicit constant a b .
(For example, on page 38 we showed that a 5 < 16.)

(g) Which value of b yields the smallest constant a b ? [Hint: This is a trick
question!]

20. Prove that the variant of the Blum-Floyd-Pratt-Rivest-Tarjan Select algo¬
rithm shown in Figure 1.21, which uses an extra layer of small medians to
choose the main pivot, runs in O(n) time.

MomomSelect(A[ 1.. n], k):
if n < 81

use brute force

else

m <—[n/3]
for i <— 1 to m

M[i ] <- MedianOf3(A[3i - 2.. 3i])
mm <— fm/3"|
for j <— 1 to mm

Mom[j ] <— MedianOf3 (M[3) — 2.. 3 j])
momom <— MoMOMSELECT(Mom[l ..mm], Lmm/2j)
r <— Partition(A[ 1 ..n], momom)
if fc < r

return MomomSelect(A[ 1.. r — 1], k )
else if k > r

return MomomSelect(A[> + 1 ..n],k — r)

else

return momom

Figure 1.21 . Selection by median of moms; see problem 20).

21 . (a) Suppose we are given two sorted arrays A[1.. n ] and B[1.. n]. Describe
an algorithm to find the median element in the union of A and B in
0(log rt) time. You can assume that the arrays contain no duplicate
elements.

(b) Suppose we are given two sorted arrays A[1 ..m] and B[1.. n] and an
integer k. Describe an algorithm to find the k til smallest element in

54

Exercises

AU B in 0(log(m + n)) time. For example, if k = 1, your algorithm
should return the smallest element of A U B.) [Hint: Use your solution
to part (a).]

r (c) Now suppose we are given three sorted arrays A[1.. n ], B[1.. n], and
C[l..n], and an integer k. Describe an algorithm to find the fcth smallest
element in A U B U C in 0(log n ) time.

(d) Finally, suppose we are given a two dimensional array A[1.. m, 1.. n ] in
which every row A[i, •] is sorted, and an integer k. Describe an algorithm
to find the k th smallest element in A as quickly as possible. How does
the running time of your algorithm depend on ml [Hint: Solve problem
16 first.]

Arithmetic

22 . In 1854 , archaeologists discovered Sumerians clay tablets, carved around
2000 BCE, that list the squares of integers up to 59. This discovery led some
scholars to conjecture that ancient Sumerians performed multiplication by
reduction to squaring, using an identity like x ■ y = (x 2 + y 2 — (x — y) 2 )/ 2 .
Unfortunately, those same scholars are silent on how the Babylonians
supposedly squared larger numbers. Four thousand years later, we can
finally rescue these Sumerian mathematicians from their lives of drudgery
through the power of recursion!

(a) Describe a variant of Karatsuba’s algorithm that squares any n-digit
number in 0 (n lg3 ) time, by reducing to squaring three [n/ 2 ]-digit
numbers. (Karatsuba actually did this in i 960 .)

(b) Describe a recursive algorithm that squares any n-digit number in
0(n log36 ) time, by reducing to squaring six [n/3]-digit numbers.

r (c) Describe a recursive algorithm that squares any n-digit number in
0(n log3S ) time, by reducing to squaring only five (n/3 + 0(l))-digit
numbers. [Hint: What is (a + b + c ) 2 + (a — b + c ) 2 ?]

23 . (a) Describe and analyze a variant of Karatsuba’s algorithm that multi¬

plies any m-digit number and any n-digit number, for any n > m, in
0 (nm lg3_1 ) time.

(b) Describe an algorithm to compute the decimal representation of 2 n in
0(n lg3 ) time, using the algorithm from part (a) as a subroutine. (The
standard algorithm that computes one digit at a time requires 0 (n 2 )
time.)

(c) Describe a divide-and-conquer algorithm to compute the decimal rep¬
resentation of an arbitrary n-bit binary number in 0(n lg3 ) time. [Hint:
Watch out for an extra log factor in the running time.]

55

i. Recursion

r (d) Suppose we can multiply two n-digit numbers in 0(M(n)) time. Describe
an algorithm to compute the decimal representation of an arbitrary n-bit
binary number in 0(M(n) log n) time. [Hint: The analysis is the hard
part; use a domain transformation.]

24 . Consider the following classical recursive algorithm for computing the

factorial n! of a non-negative integer n:

Factorial (n):
if n = 0

return 1

else

return n • Factorial(h — 1)

(a) How many multiplications does this algorithm perform?

(b) How many bits are required to write n ! in binary? Express your answer
in the form Q(J (n)), for some familiar function /(n). [Hint: (n/2) n ^ 2 <
nl < n n .]

(c) Your answer to (b) should convince you that the number of multiplications
is not a good estimate of the actual running time of Factorial. We
can multiply any Ic-digit number and any 1 -digit number in 0 (/c • l ) time
using either the lattice algorithm or duplation and mediation. What is
the running time of Factorial if we use this multiplication algorithm as
a subroutine?

(d) The following recursive algorithm also computes the factorial function,
but using a different grouping of the multiplications:

FallingCr, m): ((Compute n!/(n —m)!}}

if m = 0
return 1
else if m = 1
return n

else

return FallingIh, Lm/2j) • Falling^ — Lm/2j, \m/2\)

What is the running time of Falling(u, n) if we use grade-school multi¬
plication? [Hint: As usual, ignore the floors and ceilings.]

(e) Describe and analyze a variant of Karastuba’s algorithm that multiplies
any fc-digit number and any l -digit number, for any k > l, in 0 (/c •
fig 3 - i) = o(fc • l 0 - 585 ) time.

v (f) What are the running times of Factorial(u) and Falling(u, n) if we
use the modified Karatsuba multiplication from part (e)?

25 . The greatest common divisor of two positive integer x and y, denoted

gcd(x,y), is the largest integer d such that both x/d and y /d are integers.

56

Exercises

Euclid’s Elements, written around 300BC, describes the following recursive
algorithm to compute gcd(x,y): 16

EuclidGCD(x, y):

if x=y

return x
else if x > y

return EuclidGCD(x — y, y)

else

return EuclidGCD(x, y — x)

(a) Prove that EuclidGCD correctly computes gcd(x,y). Specifically:

i. Prove that EucLiDGCD(x,y) divides both x and y.

ii. Prove that every divisor of x and y is a divisor of EuclidGCD(x, y).

(b) What is the worst-case running time of EuclidGCD (x,y), as a function
of x and y? (Assume that computing x — y requires 0(logx + logy)
time.)

(c) Prove that the following algorithm also computes gcd(x,y):

FastEuclidGCD(x, y):

if x = y

return x
else if x > y

return FastEuclidGCD(x mod y,y)

else

return FASTEucLiDGCD(x,y mod x)

(d) What is the worst-case running time of FastEuclidGCD(x, y), as a
function of x and y? (Assume that computing x mod y takes 0(logx •
logy) time.)

(e) Prove that the following algorithm also computes gcd(x,y):

BinaryGCD(x, y):

if x = y

return x

else if x and y are both even

return 2 • BiNARYGCD(x/2,y/2)
else if x is even

BiNARYGCD(x/2,y)
else if y is even

BinaryGCD(x, y/2)
else if x > y

return BinaryGCD((x — y)/2,y)

else

return BinaryGCD(x, (y — x)/2)

l6 Euclid’s algorithm is sometimes incorrectly described as the first recursive algorithm, or even
the first nontrivial algorithm, even though the Egyptian duplation and mediation algorithm — which
is both nontrivial and recursive — predates Euclid by at least 1500 years.

57

i. Recursion

(f) What is the worst-case running time of FASTEucLiDGCD(x,y), as a
function of x andy? (Assume that computings— y takes O(logx-l-logy)
time, and computing z /2 requires O(logz) time.)

Arrays

26 . Suppose you are given a 2" x 2" chessboard with one (arbitrarily chosen)
square removed. Describe and analyze an algorithm to compute a tiling of
the board by without gaps or overlaps by L-shaped tiles, each composed of 3
squares. Your input is the integer n and two n-bit integers representing the
row and column of the missing square. The output is a list of the positions
and orientations of (4 n — l)/3 tiles. Your algorithm should run in 0(4")
time. [Hint: First prove that such a tiling always exists.]

27 . You are a visitor at a political convention (or perhaps a faculty meeting) with
n delegates; each delegate is a member of exactly one political party. It is
impossible to tell which political party any delegate belongs to; in particular,
you will be summarily ejected from the convention if you ask. However, you
can determine whether any pair of delegates belong to the same party by
introducing them to each other. Members of the same political party always
greet each other with smiles and friendly handshakes; members of different
parties always greet each other with angry stares and insults . 17

(a) Suppose more than half of the delegates belong to the same political
party. Describe an efficient algorithm that identifies all members of this
majority party.

(b) Now suppose there are more than two parties, but one party has a
plurality: more people belong to that party than to any other party.
Present a practical procedure to precisely pick the people from the
plurality political party as parsimoniously as possible, presuming the
plurality party is composed of at least p people. Pretty please.

28 . Smullyan Island has three types of inhabitants: knights always speak the
truth; knaves always lie; and normals sometimes speak the truth and
sometimes don’t. Everyone on the island knows everyone else’s name and
type (knight, knave, or normal). You want to learn the type of every
inhabitant.

You can ask any inhabitant to tell you the type of any other inhabitant.
Specifically, if you ask “Hey X, what is Y’s type?” then X will respond as
follows:

17 Real-world politics is much messier than this simplified model, but this is a theory book!

58

Exercises

• If X is a knight, then X will respond with T’s correct type.

• If X is a knave, then X could respond with either of the types that Y is
not.

• If X is a normal, then X could respond with any of the three types.

The inhabitants will ignore any questions not of this precise form; in
particular, you may not ask an inhabitant about their own type. Asking the
same inhabitant the same question multiple times always yields the same
answer, so there’s no point in asking any question more than once.

(a) Suppose you know that a strict majority of inhabitants are knights.
Describe an efficient algorithm to identify the type of every inhabitant.

(b) Prove that if at most half the inhabitants are knights, it is impossible to
determine the type of every inhabitant.

29 . Most graphics hardware includes support for a low-level operation called blit,
or block transfer, which quickly copies a rectangular chunk of a pixel map
(a two-dimensional array of pixel values) from one location to another. This
is a two-dimensional version of the standard C library function memcpy().

Suppose we want to rotate an nxn pixel map 90° clockwise. One way to
do this, at least when n is a power of two, is to split the pixel map into four
n/2 x n/2 blocks, move each block to its proper position using a sequence of
five blits, and then recursively rotate each block. (Why five? For the same
reason the Tower of Hanoi puzzle needs a third peg.) Alternately, we could
first recursively rotate the blocks and then blit them into place.

Figure 1.22. Two algorithms for rotating a pixel map.

(a) Prove that both versions of the algorithm are correct when n is a power
of 2 .

(b) Exactly how many blits does the algorithm perform when n is a power
of 2 ?

(c) Describe how to modify the algorithm so that it works for arbitrary n,
not just powers of 2. How many blits does your modified algorithm
perform?

(d) What is your algorithm’s running time if a k x k blit takes 0(/c 2 ) time?

(e) What if a k x k blit takes only O(k) time?

30 . An array A[0.. n — 1] of n distinct numbers is bitonic if there are unique

indices i and j such that A[(i — 1) mod n] < A[i] > A[(i + 1) mod n] and

59

i. Recursion

Figure 1.23. The first rotation algorithm (blit then recurse) in action. (See Image Credits at the end of
the book.)

A[(j — 1) mod n] > A[j ] < A[(j + 1) mod n], In other words, a bitonic
sequence either consists of an increasing sequence followed by a decreasing
sequence, or can be circularly shifted to become so. For example,

is bitonic, but
is not bitonic.

Describe and analyze an algorithm to find the smallest element in an n-
element bitonic array in 0(log n) time. You may assume that the numbers
in the input array are distinct.

31 . Suppose we are given an arrayA[1.. n] of n distinct integers, which could be
positive, negative, or zero, sorted in increasing order so thatA[l] < A[2] <
••• <A[n],

(a) Describe a fast algorithm that either computes an index i such that
A[i] = i or correctly reports that no such index exists.

(b) Suppose we know in advance that A[l] > 0. Describe an even faster
algorithm that either computes an index i such that A[i] = i or correctly
reports that no such index exists. [Hint: This is really easy.]

32 . Suppose we are given an array A[l..n] with the special property that
A[l] > A[2] and A[n — 1] < A[n], We say that an element A[x] is a local
minimum if it is less than or equal to both its neighbors, or more formally,
if A[x — 1] > A[x] and A[x] < A[x + 1]. For example, there are six local
minima in the following array:

4

6

9

8

7

5

1

2

3

3

6

9

8

7

5

1

2

4

60

9

7

7

2

1

3

7

5

4

7

3

3

4

8

6

9

A A AAA A

Exercises

We can obviously find a local minimum in 0(n) time by scanning through
the array. Describe and analyze an algorithm that finds a local minimum in
O(logn) time. [Hint: With the given boundary conditions, the array must
have at least one local minimum. Why?]

33 . Suppose you are given a sorted array of n distinct numbers that has been
rotated k steps, for some unknown integer k between 1 and n — 1. That is,
you are given an array A[1.. n] such that some prefix A[1.. k] is sorted in
increasing order, the corresponding suffix A[k + 1.. n] is sorted in increasing
order, and A[n] <A[1],

For example, you might be given the following 16 -element array (where
k = 10 ):

9

13

16

18

19

23

28

31

37

42

1

3

4

5

7

8

(a) Describe and analyze an algorithm to compute the unknown integer k.

(b) Describe and analyze an algorithm to determine if the given array
contains a given number x.

34 . At the end of the second act of the action blockbuster Fast and Impossible
XIII | : The Last Guardians of Expendable Justice Reloaded, the villainous
Dr. Metaphor hypnotizes the entire Hero League/Force/Squad, arranges
them in a long line at the edge of a cliff, and instructs each hero to shoot
the closest taller heroes to their left and right, at a prearranged signal.

Suppose we are given the heights of all n heroes, in order from left
to right, in an array Ht[l.. n], (To avoid salary arguments, the producers
insisted that no two heroes have the same height.) Then we can compute
the Left and Right targets of each hero in 0(n 2 ) time using the following
brute-force algorithm.

WHOTARGETSWHOM(Htr 1.. n~|):

for j <— 1 to n

((Find the left target L [j ] for hero j ))

L[j] <— None
for i <— 1 to j — 1
if Ht[i] >Ht[j]

L[j ] - i

((Find the right target R[j] for hero j))

R[j] <— None
for k <— n down to j +1
if Ht[k]>Ht[j]

R[jl «- k

return L[1 ..n], R[1 ..n]

61

i. Recursion

(a) Describe a divide-and-conquer algorithm that computes the output of
WhoTargetsWhom in O(nlogn) time.

(b) Prove that at least [n/2j of the n heroes are targets. That is, prove that
the output arrays i?[0.. n — 1] and L[0.. n — 1] contain at least \_n/2\
distinct values (other than None).

(c) Alas, Dr. Metaphor’s diabolical plan is successful. At the prearranged
signal, all the heroes simultaneously shoot their targets, and all targets
fall over the cliff, apparently dead. Metaphor repeats his dastardly
experiment over and over; after each massacre, he forces the remaining
heroes to choose new targets, following the same algorithm, and then
shoot their targets at the next signal. Eventually, only the shortest
member of the Hero Crew/Alliance/Posse is left alive. 18

Describe and analyze an algorithm to compute the number of rounds
before Dr. Metaphor’s deadly process finally ends. For full credit, your
algorithm should run in O(n) time.

35 . You are a contestant on the hit game show “Beat Your Neighbors!” You are
presented with an m x n grid of boxes, each containing a unique number. It
costs $100 to open a box. Your goal is to find a box whose number is larger
than its neighbors in the grid (above, below, left, and right). If you spend
less money than any of your opponents, you win a week-long trip for two to
Las Vegas and a year’s supply of Rice-A-Roni™, to which you are hopelessly
addicted.

(a) Suppose m = 1. Describe an algorithm that finds a number that is bigger
than either of its neighbors. How many boxes does your algorithm open
in the worst case?

T (b) Suppose m = n. Describe an algorithm that finds a number that is bigger
than any of its neighbors. How many boxes does your algorithm open in
the worst case?

* v (c) Prove that your solution to part (b) is optimal up to a constant factor.

36 . (a) Let n = 2 f — 1 for some positive integer l. Suppose someone claims to

hold an unsorted array A[1.. n] of distinct /-bit strings; thus, exactly one
£-bit string does not appear in A. Suppose further that the only way we
can access A is by calling the function FETCHBrr(i, _/), which returns the
jth bit of the string A[i] in 0(1) time. Describe an algorithm to find the
missing string in A using only 0(n) calls to FetchBit.

l8 In the thrilling final act, Retcon the Squirrel, the last surviving member of the Hero
Team/Group/Society, saves everyone by traveling back in time and retroactively replacing the
other n — 1 heroes with lifelike balloon sculptures. So, yeah, basically it’s Avengers: Endgame.

62

Exercises

r (b) Now suppose n = 2 ( — k for some positive integers k and l, and again we
are given an array A[1.. n] of distinct £-bit strings. Describe an algorithm
to find the k strings that are missing from A using only 0 (n log fc) calls
to FetchBit.

Trees

37 . For this problem, a subtree of a binary tree means any connected subgraph.
A binary tree is complete if every internal node has two children, and every
leaf has exactly the same depth. Describe and analyze a recursive algorithm
to compute the largest complete subtree of a given binary tree. Your algorithm
should return both the root and the depth of this subtree. See Figure 1.24
for an example.

Figure 1.24. The largest complete subtree of this binary tree has depth 3.

38 . Let T be a binary tree with n vertices. Deleting any vertex v splits T into at
most three subtrees, containing the left child of v (if any), the right child
of v (if any), and the parent of v (if any). We call v a central vertex if
each of these smaller trees has at most n/2 vertices. See Figure 1.25 for an
example.

Describe and analyze an algorithm to find a central vertex in an arbitrary
given binary tree. [Hint: First prove that every tree has a central vertex.]

34 14

Figure 1.25. Deleting a central vertex in a 34-node binary tree, leaving subtrees with 14,7, and 12 nodes.

63

i. Recursion

39 . (a) Professor George O’Jungle has a 27 -node binary tree, in which every

node is labeled with a unique letter of the Roman alphabet or the
character &. Preorder and postorder traversals of the tree visit the nodes
in the following order:

• Preorder: IQJHLEMVOTSBRGYZKCA&FPNUDWX

• Postorder: HEMLJVQSGYRZBTCPUDNFW&XAKOI
Draw George’s binary tree.

(b) Recall that a binary tree is full if every non-leaf node has exactly two
children.

i. Describe and analyze a recursive algorithm to reconstruct an arbitrary
full binary tree, given its preorder and postorder node sequences as
input.

ii. Prove that there is no algorithm to reconstruct an arbitrary binary
tree from its preorder and postorder node sequences.

(c) Describe and analyze a recursive algorithm to reconstruct an arbitrary
binary tree, given its preorder and inorder node sequences as input.

(d) Describe and analyze a recursive algorithm to reconstruct an arbitrary
binary search tree, given only its preorder node sequence.

r (e) Describe and analyze a recursive algorithm to reconstruct an arbitrary
binary search tree, given only its preorder node sequence, in 0(n ) time.
In parts (b)-(e), assume that all keys are distinct and that the input is
consistent with at least one binary tree.

40 . Suppose we have n points scattered inside a two-dimensional box. A kd-
tree 19 recursively subdivides the points as follows. If the box contains no
points in its interior, we are done. Otherwise, we split the box into two
smaller boxes with a vertical line, through a median point inside the box
( not on its boundary), partitioning the points as evenly as possible. Then
we recursively build a kd-tree for the points in each of the two smaller
boxes, after rotating them go degrees. Thus, we alternate between splitting
vertically and splitting horizontally at each level of recursion. The final
empty boxes are called cells.

19 The term “kd-tree” (pronounced “kay dee tree”) was originally an abbreviation for “k-
dimensional tree”, but modern usage ignores this etymology, in part because nobody in their
right mind would ever use the letter k to denote dimension instead of the obviously superior d.
Etymological consistency would require calling the data structure in this problem a “2d-tree”
(or perhaps a “2-d tree”), but the standard nomenclature is now “two-dimensional kd-tree”.
See also: B-tree (maybe), alpha shape, beta skeleton, epsilon net, Potomac River, Mississippi
River, Lake Michigan, Lake Tahoe, Manhattan Island, the La Brea Tar Pits, Sahara Desert, Mount
Kilimanjaro, South Vietnam, East Timor, the Milky Way Galaxy, the City of Townsville, and
self-driving automobiles.

64

Exercises

Figure 1.26. A kd-tree for 15 points. The dashed line crosses the four shaded cells.

(a) How many cells are there, as a function of n ? Prove your answer is
correct.

(b) In the worst case, exactly how many cells can a horizontal line cross, as
a function of n ? Prove your answer is correct. Assume that n = 2 k — 1
for some integer k. [Hint: There is more than one function f such that
/(16) = 4J

(c) Suppose we are given n points stored in a kd-tree. Describe and analyze
an algorithm that counts the number of points above a horizontal line
(such as the dashed line in the figure) as quickly as possible. [Hint: Use
part (b).]

(d) Describe an analyze an efficient algorithm that counts, given a kd-tree
containing n points, the number of points that lie inside a rectangle R
with horizontal and vertical sides. [Hint: Use part (c).]

* 41 . Bob Ratenbur, a new student in CS 225 , is trying to write code to perform
preorder, inorder, and postorder traversals of binary trees. Bob sort-of
understands the basic idea behind the traversal algorithms, but whenever
he actually tries to implement them, he keeps mixing up the recursive calls.
Five minutes before the deadline, Bob frantically submits code with the
following structure:

PreOrder(v):

InOrder(v):

PostOrder(v):

if v = Null

if v = Null

if v = Null

return

return

return

else

else

else

print ZabeZfv)

^BORDER(iett(v))

■ORDER(Ze/t(v))

^■ORDER(ie/t(v))

print ZabeZ(v)

^MORDERfrigh t(v))

^MORDER(rigfrt(v))

^BORDER(right(v))

print ZabeZ(v)

Each in this pseudocode hides one of the prefixes Pre, In, or Post.
Moreover, each of the following function calls appears exactly once in Bob’s
submitted code:

65

i. Recursion

PREORDER(7e/t(v)) PREORDER(rigfrt(v))

lNORDER(Ze/t(v)) lNORDER(rigfrt(v))

POST0RDER(Ze/t(v)) PosTORDER(right(v))

Thus, there are precisely 36 possibilities for Bob’s code. Unfortunately, Bob
accidentally deleted his source code after submitting the executable, so
neither you nor he knows which functions were called where.

Now suppose you are given the output of Bob’s traversal algorithms,
executed on some unknown binary tree T. Bob’s output has been helpfully
parsed into three arrays Pre[l..n], In[l..n], and Post[l..n]. You may
assume that these traversal sequences are consistent with exactly one binary
tree T; in particular, the vertex labels of the unknown tree T are distinct,
and every internal node in T has exactly two children.

(a) Describe an algorithm to reconstruct the unknown tree T from the given
traversal sequences.

(b) Describe an algorithm that either reconstructs Bob’s code from the given
traversal sequences, or correctly reports that the traversal sequences are
consistent with more than one set of algorithms.

For example, given the input

Pre[l..n] = [HAECBIFGD]

In[l.. n] = [A H D C E I F B G]

Post[l..n] = [A E I B F C D G H]

your first algorithm should return the following tree:

and your second algorithm should reconstruct the following code:

PreOrder(v):

InOrder(v):

PostOrder(V):

if v = Null

if v = Null

if v = Null

return

return

return

else

else

else

print ZabeZ(v)

PoSTORDER(Ze/t(v))

lNORDER(Ze/t(v))

PREORDER(Ze/t(v))

print ZabeZ(v)

lNORDER(right(v))

POST0RDER(rig/lt(v))

PREORDER(right(v))

print Zabei(v)

r 4 2 . Let T be a binary tree whose nodes store distinct numerical values. Recall
that T is a binary search tree if and only if either ( 1 ) T is empty, or ( 2 ) T
satisfies the following recursive conditions:

66

Exercises

• The left subtree of T is a binary search tree.

• All values in the left subtree are smaller than the value at the root.

• The right subtree of T is a binary search tree.

• All values in the right subtree are larger than the value at the root.
Consider the following pair of operations on binary trees:

• Rotate an arbitrary node upward . 20

• Swap the left and right subtrees of an arbitrary node.

In both of these operations, some, all, or none of the subtrees A, B, and C
could be empty.

(a) Describe an algorithm to transform an arbitrary n-node binary tree
with distinct node values into a binary search tree, using at most 0 (n 2 )
rotations and swaps. Figure 1.27 shows a sequence of eight operations
that transforms a five-node binary tree into a binary search tree.

Figure 1.27. "Sorting" a binary tree: rotate 2, rotate 2, swap 3, rotate 3, rotate 4, swap 3, rotate 2, swap 4.

Your algorithm is not allowed to directly modify parent or child
pointers, create new nodes, or delete old nodes; the only way to modify
the tree is through rotations and swaps.

On the other hand, you may compute anything you like for free, as
long as that computation does not modify the tree; the running time of
your algorithm is defined to be the number of rotations and swaps that it
performs.

r (b) Describe an algorithm to transform an arbitrary n-node binary tree into
a binary search tree, using at most 0 (n log n) rotations and swaps.

“Rotations preserve the inorder sequence of nodes in a binary tree. Partly for this reason,
rotations are used to maintain several types of balanced binary search trees, including AVL trees,
red-black trees, splay trees, scapegoat trees, and treaps. See http://algorithms.wtf for lecture
notes on all of these data structures.

67

i. Recursion

(c) Prove that any n-node binary search tree can be transformed into any
other binary search tree with the same node values, using only O(n)
rotations (and no swaps).

V(d) Open problem: Either describe an algorithm to transform an arbitrary
n-node binary tree into a binary search tree using only O(n) rotations
and swaps, or prove that no such algorithm is possible. [Hint: I don’t
think it’s possible.]

68

Where, however, the ambiguity cannot be cleared up, either by the rule of faith or by
the context, there is nothing to hinder us to point the sentence according to any
method we choose of those that suggest themselves.

— Augustine of Hippo, De doctrina Christiana (397CE)
Translated by Marcus Dods (1892)

I dropped my dinner, and ran back to the laboratory. There, in my excitement,

I tasted the contents of every beaker and evaporating dish on the table. Luckily for
me, none contained any corrosive or poisonous liquid.

— Constantine Fahlberg on his discovery of saccharin,
Scientific American (1886)

The greatest challenge to any thinker is stating the problem
in a way that will allow a solution.

- attributed to Bertrand Arthur William Russell
When you come to a fork in the road, take it.

— Yogi Berra (giving directions to his house)

Backtracking

This chapter describes another important recursive strategy called backtracking.
A backtracking algorithm tries to construct a solution to a computational problem
incrementally, one small piece at a time. Whenever the algorithm needs to
decide between multiple alternatives to the next component of the solution, it
recursively evaluates every alternative and then chooses the best one.

2.1 N Queens

The prototypical backtracking problem is the classical n Queens Problem, first
proposed by German chess enthusiast Max Bezzel in 1848 (under his pseudonym
“Schachfreund”) for the standard 8x8 board and by Francois-Joscph Eustache
Lionnet in 1869 for the more general n x n board. The problem is to place n
queens on an n x n chessboard, so that no two queens are attacking each other.

69

2. Backtracking

For readers not familiar with the rules of chess, this means that no two queens
are in the same row, the same column, or the same diagonal.

Figure 2.1. One solution to the 8 queens problem, represented by the array [ 4 , 7 , 3 ,8, 2 , 5 , 1 , 6 ]

In a letter written to his friend Heinrich Schumacher in 1850, the eminent
mathematician Carl Friedrich Gauss wrote that one could easily confirm Franz
Nauck’s claim that the Eight Queens problem has 92 solutions by trial and
error in a few hours. (“ Schwer ist es ilbrigens nicht, durch ein methodisches
Tatonnieren sich diese Gewifsheit zu verschaffen, wenn man eine oder ein paar
Stunden daran wenden will”) His description Tatonnieren comes from the French
tattoner, meaning to feel, grope, or fumble around blindly, as if in the dark.
Unfortunately, Gauss did not describe the mechanical groping method he had in
mind, but he did observe that any solution can be represented by a permutation
of the integers 1 through 8 satisfying a few simple arithmetic properties.

Following Gauss, let’s represent possible solutions to the n-queens problem
using an array Q[ 1 .. n], where Q[t] indicates which square in row i contains
a queen. Then we can find solutions using the following recursive strategy,
described in 1882 by the French recreational mathematician Edouard Lucas,
who attributed the method to Emmanuel Laquiere. We place queens on the
board one row at a time, starting with the top row. To place the rth queen, we
methodically try all n squares in row r from left to right in a simple for loop.
If a particular square is attacked by an earlier queen, we ignore that square;
otherwise, we tentatively place a queen on that square and recursively grope for
consistent placements of the queens in later rows.

Figure 2.2 shows the resulting algorithm, which recursively enumerates all
complete n-queens solutions that are consistent with a given partial solution.
The input parameter r is the index of the first empty row; the prefix Q[ 1 .. r — 1 ]
contains the position of the first r — 1 queens. In particular, to compute all n-
queens solutions with no restrictions, we would call PlaceQueens(Q[1 .. n], 1).
The outer for-loop considers all possible placements of a queen on row r; the
inner for-loop checks whether a candidate placement of row r is consistent with
the queens that are already on the first r — 1 rows.

70

2.1. N Queens

PlaceQueens(Q|~ 1.. nl, r):

if r = n + 1
print Q

else

for j <— 1 to n
legal <— True
for i <— 1 to r — 1

if (Q[i] = j) or (Q[i] = ; + r - i) or (Q[i] = j - r + i)
legal <— False

if legal

PlaceQueens(Q[1 .. n], r + 1) ((Recursion!))
Figure 2.2. Laquiere’s backtracking algorithm for the n-queens problem.

The execution of PlaceQueens can be illustrated using a recursion tree.
Each node in this tree corresponds to recursive subproblem, and thus to a
legal partial solution; in particular, the root corresponds to the empty board
(with r = 0 ). Edges in the recursion tree correspond to recursive calls. Leaves
correspond to partial solutions that cannot be further extended, either because
there is already a queen on every row, or because every position in the next
empty row is attacked by an existing queen. The backtracking search for
complete solutions is equivalent to a depth-first search of this tree.

Figure 2.3. The complete recursion tree of Laquiere's algorithm for the 4 queens problem.

71

2. Backtracking

2.2 Game Trees

Consider the following simple two-player game 1 played on an n x n square grid
with a border of squares; let’s call the players Horace Fahlberg-Remsen and
Vera Rebaudi. 2 Each player has n tokens that they move across the board from
one side to the other. Horace’s tokens start in the left border, one in each row,
and move horizontally to the right; symmetrically, Vera’s tokens start in the
top border, one in each column, and move vertically downward. The players
alternate turns. In each of his turns, Horace either moves one of his tokens
one step to the right into an empty square, or jumps one of his tokens over
exactly one of Vera’s tokens into an empty square two steps to the right. If no
legal moves or jumps are available, Horace simply passes. Similarly, Vera either
moves or jumps one of her tokens downward in each of her turns, unless no
moves or jumps are possible. The first player to move all their tokens off the
edge of the board wins. (It’s not hard to prove that as long as there are tokens
on the board, at least one player can legally move, so someone eventually wins.)

©

©

©

0

0

©

©

©

©

◄

0

0

©

©

©

0

0

©

©

©

©

0

©

0

©

©

0

©

©

0

©

©

©

©

©

0

©

©

©

©

©

0

0

©

©

©

<

0

©

©

0

©

0

©

0

©

©

©

0

©

©

©

Figure 2.4. Vera wins the 3 x 3 fake-sugar-packet game.

T don’t know what this game is called, or even if I’m remembering the rules correctly; I
learned it (or something like it) from Lenny Pitt, who recommended playing it with fake-sugar
packets at restaurants.

2 Constantin Fahlberg and Ira Remsen synthesized saccharin for the first time in 1878, while
Fahlberg was a postdoc in Remsen’s lab investigating coal tar derivatives. In 1900, Ovidio Rebaudi
published the first chemical analysis of ka’a he’i, a medicinal plant cultivated by the Guarani for
more than 1500 years, now more commonly known as Stevia rebaudiana.

72

2.2. Game Trees

Unless you’ve seen this game before 3 , you probably don’t have any idea how
to play it well. Nevertheless, there is a relatively simple backtracking algorithm
that can play this game—or any two-player game without randomness or hidden
information— perfectly. That is, if we drop you into the middle of a game, and it
is possible to win against another perfect player, the algorithm will tell you how
to win.

A state of the game consists of the locations of all the pieces and the identity
of the current player. These states can be connected into a game tree, which has
an edge from state x to state y if and only if the current player in state x can
legally move to state y. The root of the game tree is the initial position of the
game, and every path from the root to a leaf is a complete game.

CO

CO

CO

0

0

0

CO

CO

CO

CO

CO

CO

0

0

CO

0

CO

0

CO

C-J

C-)

0

C-)

c-)

CO

CO

CO

CO

CO

CO

0

(0

0

(0

0

(0

0

0

0

C-)

0

0

0 ®

®©

i /I' /I

ffi®

Figure 2.5. The first two levels of the fake-sugar-packet game tree.

To navigate through this game tree, we recursively define a game state to
be good or bad as follows:

• A game state is good if either the current player has already won, or if the
current player can move to a bad state for the opposing player.

• A game state is bad if either the current player has already lost, or if every
available move leads to a good state for the opposing player.

Equivalently, a non-leaf node in the game tree is good if it has at least one bad
child, and a non-leaf node is bad if all its children are good. By induction, any
player that finds the game in a good state on their turn can win the game, even
if their opponent plays perfectly; on the other hand, starting from a bad state, a
player can win only if their opponent makes a mistake.

This recursive definition immediately suggests the following recursive back¬
tracking algorithm to determine whether a given game state is good or bad. At
its core, this algorithm is just a depth-first search of the game tree; equivalently,
the game tree is the recursion tree of the algorithm. A simple modification of

3 If you have, please tell me where!

73

2. Backtracking

this algorithm finds a good move (or even all possible good moves) if the input
is a good game state.

PlayAnyGame(X, player ):
if player has already won in state X
return Good

if player has already lost in state X
return Bad

for all legal moves X Y

if PlayAnyGame(Y, -i player ) = Bad

return Good ((X Y is a good move))
return Bad ((There are no good moves))

All game-playing programs are ultimately based on this simple backtracking
strategy. However, since most games have an enormous number of states, it is
not possible to traverse the entire game tree in practice. Instead, game programs
employ other heuristics 4 to prune the game tree, by ignoring states that are
obviously (or “obviously”) good or bad, or at least better or worse than other
states, and/or by cutting off the tree at a certain depth (or ply ) and using a
more efficient heuristic to evaluate the leaves.

2.3 Subset Sum

Let’s consider a more complicated problem, called SubsetSum: Given a set A
of positive integers and target integer T, is there a subset of elements in X that
add up to T? Notice that there can be more than one such subset. For example,
if A = {8,6,7,5,3,10,9} and T = 15, the answer is True, because the subsets
{8,7} and {7,5,3} and {6,9} and {5,10} all sum to 15. On the other hand, if
X = {11,6,5,1,7,13,12} and T = 15, the answer is False.

There are two trivial cases. If the target value T is zero, then we can

immediately return True, because empty set is a subset of every set X, and the

elements of the empty set add up to zero. 5 On the other hand, if T <0, or if
T / 0 but the set X is empty, then we can immediately return False.

For the general case, consider an arbitrary element x €X. (We’ve already
handled the case where X is empty.) There is a subset ofX that sums to T if
and only if one of the following statements is true:

• There is a subset of X that includes x and whose sum is T.

• There is a subset of X that excludes x and whose sum is T.

74

4 A heuristic is an algorithm that doesn’t work. (Except in practice. Sometimes. Maybe.)

5 ... because what else could they add up to?

2 . 3 - Subset Sum

In the first case, there must be a subset of X \ {x} that sums to T — x; in the
second case, there must be a subset of X \ {x} that sums to T. So we can solve
SubsetSum(X, T) by reducing it to two simpler instances: SubsetSum(X \ {x},
T — x) and SubsetSum(X \ {x}, T ). The resulting recursive algorithm is shown
below.

((Does any subset ofX sum to T ?))

SubsetSum(X, T):

if T = 0

return True

else if T < 0 or X = 0

return False

else

x <— any element of A

with <— SubsetSum(X \ {x}, T — x)

((Recurse!))

wout <— SubsetSum(X \ {x}, T)
return (with V wout )

((Recurse!))

Correctness

Proving this algorithm correct is a straightforward exercise in induction. If
T = 0, then the elements of the empty subset sum to T, so True is the correct
output. Otherwise, if T is negative or the set X is empty, then no subset of X
sums to T, so False is the correct output. Otherwise, if there is a subset that
sums to T, then either it contains X[n] or it doesn’t, and the Recursion Fairy
correctly checks for each of those possibilities. Done.

Analysis

In order to analyze the algorithm, we have to describe a few implementation
details more precisely. To begin, let’s assume that the input set X is given as an
array X[1.. n\.

The previous recursive algorithm allows us to choose any element x e X in
the main recursive case. Purely for the sake of efficiency, it is helpful to choose
an element x such that the remaining subset X \ {x} has a concise representation,
which can be computed quickly, so that we pay minimal overhead making the
recursive calls. Specifically, we will let x be the last element X[n]; then the
subsetX \ {x} is stored in the prefixX[l.. n— 1]. Passing a complete copy of
this prefix to the recursive calls would take too long—we need 0(n) time just
to make the copy—so instead, we push only two values: a reference to the
array (or its starting address) and the length of the prefix. (Alternatively, we
could avoid passing a reference to X to every recursive call by making X a global
variable.)

75

2. Backtracking

((Does any subset of X[1 ..i] sum to T?))

SubsetSum(X,i, T):

if T = 0

return True

else if T < 0 or i = 0

return False

else

with <— SubsetSum(X,i — 1, T — X[i])

((Recurse!))

wout <— SubsetSum(X, i — 1, T)

((Recurse!))

return (with V wout )

With these implementation choices, the running time T (n) of our algorithm
satisfies the recurrence T(n) < 2 T(n — 1) + 0(1). From its resemblance to the
Tower of Hanoi recurrence, we can guess the solution T(n) = 0(2 n ); verifying
this solution is another easy induction exercise. (We can also derive the solution
directly, using recursion trees.) In the worst case—for example, when T is larger
than the sum of all elements of X—the recursion tree for this algorithm is a
complete binary tree with depth n, and the algorithm considers all 2 n subsets
of X.

Variants

With only minor changes, we can solve several variants of SubsetSum. For
example, Figure 2.6 shows an algorithm that actually constructs a subset of X
that sums to T, if one exists, or returns the error value None if no such subset
exists; this algorithm uses exactly the same recursive strategy as our earlier
decision algorithms. This algorithm also runs in 0(2") time; the analysis is
simplest if we assume a set data structure that allows us to insert a single
element in 0 ( 1 ) time (for example, a singly-linked list), but in fact the running
time is still 0(n) even if adding an element to Y in the second-to-last time
requires 0(|T[) time. Similar variants allow us to count subsets that sum to a
particular value, or choose the best subset (according to some other criterion)
that sums to a particular value.

Most other problems that are solved by backtracking have this property:
the same recursive strategy can be used to solve many different variants of the
same problem. For example, it is easy to modify the recursive strategy described
in the previous section, which determines whether a given game position is
good or bad, to instead return a good move, or a list of all good moves. For
this reason, when we design backtracking algorithms, we should aim for the
simplest possible variant of the problem, computing a number or even a single
boolean instead of more complex information or structure.

76

2 . 4 - The General Pattern

(( Return a subset ofX[ 1.. i] that sums to T))

((or None if no such subset exists))

ConstructSubset(X, i,T ):
if T = 0

return 0
if T < 0 or n = 0
return None

Y «— ConstructSubset(X,i — 1, T)
if Y 7 ^ None
return 7

7 <— ConstructSubset(X, i — 1, T — X[i])
if 7 7 ^ None

return Y u {X[i]}
return None

Figure 2.6. A recursive backtracking algorithm for the construction version of SubsetSum.

2.4 The General Pattern

Backtracking algorithms are commonly used to make a sequence of decisions, with
the goal of building a recursively defined structure satisfying certain constraints.
Often (but not always) this goal structure is itself a sequence. For example:

• In the n-queens problem, the goal is a sequence of queen positions, one in
each row, such that no two queens attack each other. For each row, the
algorithm decides where to place the queen.

• In the game tree problem, the goal is a sequence of legal moves, such that
each move is as good as possible for the player making it. For each game
state, the algorithm decides the best possible next move.

• In the subset sum problem, the goal is a sequence of input elements that have
a particular sum. For each input element, the algorithm decides whether to
include it in the output sequence or not.

(Hang on, why is the goal of subset sum finding a sequence ? That was a
deliberate design decision. We imposed a convenient ordering on the input
set—by representing it using an array as opposed to some other more amorphous
data structure—that we can exploit in our recursive algorithm.)

In each recursive call to the backtracking algorithm, we need to make exactly
one decision, and our choice must be consistent with all previous decisions.
Thus, each recursive call requires not only the portion of the input data we have
not yet processed, but also a suitable summary of the decisions we have already
made. For the sake of efficiency, the summary of past decisions should be as
small as possible. For example:

77

2. Backtracking

• For the n-queens problem, we must pass in not only the number of empty
rows, but the positions of all previously placed queens. Here, unfortunately,
we must remember our past decisions in complete detail.

• For the game tree problem, we only need to pass in the current state of the
game, including the identity of the next player. We don’t need to remember
anything about our past decisions, because who wins from a given game
state does not depend on the moves that created that state. 6

• For the subset sum problem, we need to pass in both the remaining available
integers and the remaining target value, which is the original target value
minus the sum of the previously chosen elements. Precisely which elements
were previously chosen is unimportant.

When we design new recursive backtracking algorithms, we must figure out in
advance what information we will need about past decisions in the middle of the
algorithm. If this information is nontrivial, our recursive algorithm must solve a
more general problem than the one we were originally asked to solve. (We’ve
seen this kind of generalization before: To find the median of an unsorted array
in linear time, we derived an algorithm to find the kth smallest element for
arbitrary k .)

Finally, once we’ve figured out what recursive problem we really need to
solve, we solve that problem by recursive brute force: Try all possibilities for
the next decision that are consistent with past decisions, and let the Recursion
Fairy worry about the rest. No being clever here. No skipping “obviously” stupid
choices. Try everything. You can make the algorithm faster later.

2.5 String Segmentation (Interpunctio Verborum)

Suppose you are given a string of letters representing text in some foreign
language, but without any spaces or punctuation, and you want to break this
string into its individual constituent words. For example, you might be given
the following passage from Cicero’s famous oration in defense of Lucius Licinius
Murena in 62 bce , in the standard scriptio continua of classical Latin: 7

6 Many games violate this independence condition. For example, the standard rules of both
chess and checkers allow a player to declare a draw if the same arrangement of pieces occurs
three times, and the Chinese rules for go simply forbid repeating any earlier arrangement of
stones. Thus, for these games, a game state formally includes the entire history of previous
moves.

qn-fact-most-classical-Latin-manuscripts-separated-words-with-small-dots-called-interpuncts.
Interpunctuation all but disappeared by the 3rd century in favor of scriptio continua. Empty
spaces between words were introduced by Irish monks in the 8th century and slowly spread
across Europe over the next several centuries. Scriptio continua survives in early 21st-century
English in the form of URLs and hashtags. #octotherps4lyfe

78

2.5. String Segmentation (Interpunctio Verborum)

PRIMVSDIGNITASINTAMTENVI SCI ENTIANONPOTEST
ESSERESENIMSVNTPARVAEPROPEINSINGVLISLITTERIS
ATQVEINTERPVNCTIONIBUSVERBORVMOCCVPATAE

A fluent Latin reader would parse this string (in modern orthography) as
Primus dignitas in tam tenui scientia non potest esse; res enim sunt parvae,
prope in singulis litteris atque interpunctionibus verborum occupatae. 8 Text
segmentation is not only a problem in classical Latin and Greek, but in several
modern languages and scripts including Balinese, Burmese, Chinese, Japanese,
Javanese, Khmer, Lao, Thai, Tibetan, and Vietnamese. Similar problems arise in
segmenting unpunctuated English text into sentences, 9 segmenting text into
lines for typesetting, speech and handwriting recognition, curve simplification,
and several types of time-series analysis. For purposes of illustration, I’ll stick to
segmenting sequences of letters in the modern English alphabet into modern
English words.

Of course, some strings can be segmented in several different ways; for
example, BOTHEARTHANDSATURNSPIN can be decomposed into English words
as either BOTH-EARTH-AND-SATURN-SPIN or BOT-HEART-HANDS-AT-URNS-PIN,
among many other possibilities. For now, let’s consider an extremely simple
segmentation problem: Given a string of characters, can it be segmented into
English words at all ?

To make the problem concrete (and language-agnostic), let’s assume we
have access to a subroutine IsWord(w) that takes a string w as input, and
returns True if w is a “word”, or False if w is not a “word”. For example, if
we are trying to decompose the input string into palindromes, then a “word”
is a synonym for “palindrome”, and therefore IsWord(ROTATOR) = True but
IsWord(PALINDROME) = False.

Just like the subset sum problem, the input structure is a sequence, this time
containing letters instead of numbers, so it is natural to consider a decision
process that consumes the input characters in order from left to right. Similarly,
the output structure is a sequence of words, so it is natural to consider a process
that produces the output words in order from left to right. Thus, jumping into
the middle of the segmentation process, we might imagine the following picture:

BLUE

STEM

UNIT

ROBOT

HEARTHANDSATURNSPIN

8 Loosely translated: “First of all, dignity in such paltry knowledge is impossible; this is trivial
stuff, mostly concerned with individual letters and the placement of points between words.”
Cicero was openly mocking the legal expertise of his friend(!) and noted jurist Servius Sulpicius
Rufus, who had accused Murena of bribery, after Murena defeated Rufus in election for consul.
Murena was acquitted, thanks in part to Cicero’s acerbic defense, although he was almost certainly
guilty. #librapondo #nunquamestfidelis

9 St. Augustine’s De doctrina Christiana devotes an entire chapter to removing ambiguity from
Latin scripture by adding punctuation.

79

2. Backtracking

Here the black bar separates our past decisions—splitting the first 17 letters into
four words—from the portion of the input string that we have not yet processed.

The next stage in our imagined process is to decide where the next word in
the output sequence ends. For this specific example, there are four possibilities
for the next output word— HE, HEAR, HEART, and HEARTH. We have no idea which
of these choices, if any, is consistent with a complete segmentation of the input
string. We could be “smart” at this point and try to figure out which choices
are good, but that would require thinking ! Instead, let’s “stupidly” try every
possibility by brute force, and let the Recursion Fairy do all the real work.

• First tentatively accept HE as the next word, and let the Recursion Fairy make
the rest of the decisions.

BLUE

STEM

UNIT

ROBOT

HE

ARTHANDSATURNSPIN

• Then tentatively accept HEAR as the next word, and let the Recursion Fairy
make all remaining decisions.

BLUE

STEM

UNIT

ROBOT

HEAR

THANDSATURNSPIN

• Then tentatively accept HEART as the next word, and let the Recursion Fairy
make all remaining decisions.

BLUE

STEM

UNIT

ROBOT

HEART

HANDSATURNSPIN

• Finally, tentatively accept HEARTH as the next word, and let the Recursion
Fairy make all remaining decisions.

BLUE

STEM

UNIT

ROBOT

HEARTH

ANDSATURNSPIN

As long as the Recursion Fairy reports success at least once, we report success.
On the other hand, if the Recursion Fairy never reports success—in particular, if
the set of possible next words is empty—then we report failure.

None of our past decisions affect which choices are available now; all that
matters is the suffix of characters that we have not yet processed. In particular,
several different sequences of past decisions might have led us to the same suffix,
but they all leave us with exactly the same set of choices now.

BLUE STEM UNIT ROBOT

HEARTHANDSATURNSPIN

BLUEST

EMU

NITRO

BOT

HEARTHANDSATURNSPIN

Thus, we can simplify our picture of the recursive process by discarding everything
left of the black bar:

| HEARTHANDSATURNSPIN

80

2.5. String Segmentation (Interpunctio Verborum)

We are now left with a simple and natural backtracking strategy: Select the first
output word, and recursively segment the rest of the input string.

To get a complete recursive algorithm, we need a base case. Our recursive
strategy breaks down when we reach the end of the input string, because there
is no next word. Fortunately, the empty string has a unique segmentation into
zero words!

Putting all the pieces together, we arrive at the following simple recursive
algorithm:

SPLITTABLE(A[~ 1.. rz~|):

if n = 0

return True
for i *— 1 to n

if IsWord(A[ 1 .. i ])

if Splittable(A[i + 1.. n])
return True

return False

Index Formulation

In practice, passing arrays as input parameters to algorithm is rather slow; we
should really find a more compact way to describe our recursive subproblems.
For purposes of designing the algorithm, it’s incredibly useful to treat the original
input array as a global variable, and then reformulate the problem and the
algorithm in terms of array indices instead of explicit subarrays.

For our string segmentation problem, the argument of any recursive call
is always a suffix A[i .. n] of the original input array. So if we treat the input
array A[1.. n] as a global variable, we can reformulate our recursive problem as
follows:

Given an index i, find a segmentation of the suffix A[i .. n].

To describe our algorithm, we need two boolean functions:

• For any indices i and j, let IsWord(/, j) = True if and only if the substring
A[i .. j ] is a word. (We’re assuming this function is given to us.)

• For any index i, let Splittable(i) = True if and only if the suffix A[i .. n] can
be split into words. (This is the function we need to implement.)

For example, IsWord(1, n) = True if and only if the entire input string is a
single word, and Splittable( 1) = True if and only if the entire input string can
be segmented. Our earlier recursive strategy gives us the following recurrence:

True if i > n

Splittablefi )

< n

\J (IsWord(i,j) A Splittable{j + 1)) otherwise

J =i

8i

2. Backtracking

This is exactly the same algorithm as we saw earlier; the only thing we’ve
changed is the notation. The similarity is even more apparent if we rewrite the
recurrence in pseudocode:

((Is the suffix A[i ..n] Splittable?))

Splittable (Q:
if i > n

return True
for j *— i to n

if IsWord(i, j)

if SplittableQ' + 1)
return True

return False

Although it may look like a trivial notational difference, using index notation
instead of array notation is an important habit, not only to speed up backtracking
algorithms in practice, but for developing dynamic programming algorithms,
which we discuss in the next chapter.

^Analysis

It should come as no surprise that most backtracking algorithms have exponential
worst-case running times. Analyzing the precise running times of many of
these algorithms requires techniques that are beyond the scope of this book.
Fortunately, most of the backtracking algorithms we will encounter in this book
are only intermediate results on the way to more efficient algorithms, which
means their exact worst-case running time is not actually important. (First
make it work; then make it fast.)

But just for fun, let’s analyze the running time of our recursive algorithm
Splittable. Because we don’t know what what IsWord is doing, we can’t know
how long each call to IsWord takes, so we’re forced to analyze the running
time in terms of the number of calls to IsWord. 10 Splittable calls IsWord
on every prefix of the input string, and possibly calls itself recursively on every
suffix of the output string. Thus, the “running time” of Splittable obeys the
scary-looking recurrence

n—l

T(n)<J]T(i) + 0(n)

i=0

This really isn’t as bad as it looks, especially once you’ve seen the trick.

First, we replace the O(n) term with an explicit expression an, for some
unknown (and ultimately unimportant) constant a. Second, we conservatively

“In fact, as long as IsWord runs in polynomial time, Splittable still runs in 0 ( 2 ") time.

82

2.5. String Segmentation (Interpunctio Verborum)

assume that the algorithm actually makes every possible recursive call. 11 Then
we can transform the “full history” recurrence into a “limited history” recurrence
by subtracting the recurrence for T(n — 1), as follows:

n—1

T(n) = ^T(i) + an

i =0
n—2

Tin- 1) = ^T(i) + a(n-l)

;=o

=> T{n) — T{n — 1) = T{n— l) + a

This final recurrence simplifies to T(n) = 2T(n — 2) + a. At this point, we can
confidently guess (or derive via recursion trees) that T(n) = 0(2 n ); indeed, this
upper bound is not hard to prove by induction from the original full-history
recurrence.

Moreover, this analysis is tight. There are exactly 2 n_1 possible ways to
segment a string of length n —each input character either ends a word or doesn’t,
except the last input character, which always ends the last word. In the worst
case, our Splittable algorithm explores each of these 2 n_1 possibilities.

Variants

Now that we have the basic recursion pattern in hand, we can use it to solve
many different variants of the segmentation problem, just as we did for the
subset sum problem. Here I’ll describe just one example; more variations are
considered in the exercises. As usual, the original input to our problem is an
array A[l .. n].

If a string can be segmented in more than one sequence of words, we may
want to find the best segmentation according to some criterion; conversely, if
the input string cannot be segmented into words, we may want to compute
the best segmentation we can find, rather than merely reporting failure. To
meet both of these goals, suppose we have access to a second function Score
that takes a string w as input and returns a numerical value. For example, we
might assign higher scores to longer and/or more frequent words, and lower
negative scores to longer and/or more ridiculous non-words. Our goal is to find
a segmentation that maximizes the sum of the scores of the segments.

“This assumption is wildly conservative for English word segmentation, since most strings of
letters are not English words, but not for the similar problem of segmenting sequences of English
words into grammatically correct English sentences. Consider, for example, a sequence of n copies
of the word “buffalo”, or n copies of the work “police”, or n copies of the word “can”, for any
positive integer n. (At the Moulin Rouge, dances that are preservable in metal cylinders by other
dances have the opportunity to fire dances that happen in prison restroom trash receptacles.)

83

2. Backtracking

For any index i, let MaxScore{i ) denote the maximum score of any segmen¬
tation of the suffix A[i .. n]; we need to compute MaxScore(l). This function
satisfies the following recurrence:

This is essentially the same recurrence as the one we developed for Splittable;
the only difference is that the boolean operations V and A have been replaced
by the numerical operations max and +.

2.6 Longest Increasing Subsequence

For any sequence S, a subsequence of S is another sequence obtained from S by
deleting zero or more elements, without changing the order of the remaining
elements; the elements of the subsequence need not be contiguous in S. For
example, when you drive down a major street in any city, you drive through
a sequence of intersections with traffic lights, but you only have to stop at a
subsequence of those intersections, where the traffic lights are red. If you’re very
lucky, you never stop at all: the empty sequence is a subsequence of S. On the
other hand, if you’re very unlucky, you may have to stop at every intersection: S
is a subsequence of itself.

As another example, the strings BENT, ACKACK, SQUARING, and SUBSEQUENT
are all subsequences of the string SUBSEQUENCEBACKTRACKING, as are the empty
string and the entire string SUBSEQUENCEBACKTRACKING, but the strings QUEUE
and EQUUS and TALLYHO are not. A subsequence whose elements are contiguous
in the original sequence is called a substring; for example, MASHER and LAUGHTER
are both subsequences of MANSLAUGHTER, but only LAUGHTER is a substring.

Now suppose we are given a sequence of integers, and we need to find the
longest subsequence whose elements are in increasing order. More concretely,
the input is an integer array A[1.. n], and we need to compute the longest
possible sequence of indices 1 < ij < i 2 < • • • < it < n such that A[i k ] < A[i fc+1 ]
for all k.

One natural approach to building this longest increasing subsequence is to
decide, for each index j in order from 1 to n, whether or not to include A[j] in
the subsequence. Jumping into the middle of this decision sequence, we might
imagine the following picture:

3

1 4

1

5

9 2

6

5 3

5

897932384627

As in our segmentation example, the black bar separates our past decisions
from the portion of the input we have not yet processed. Numbers we have

2.6. Longest Increasing Subsequence

already decided to include are highlighted; numbers we have already decided to
exclude are grayed out. (Notice that the numbers we’ve decided to include are
increasing!) Our algorithm must decide whether or not to include the number
immediately after the black bar.

In this example, we cannot include 5, because then the selected numbers
would no longer be in increasing order. So let’s skip ahead to the next decision:

3

1 4

1

5

9 2

6

5 3 5

8

97932384627

Now we can include 8, but it’s not obvious whether we should. Rather than
trying to be “smart”, our backtracking algorithm will use simple brute force.

• First tentatively include the 8, and let the Recursion Fairy make the rest of
the decisions.

• Then tentatively exclude the 8, and let the Recursion Fairy make the rest of
the decisions.

Whichever choice leads to a longer increasing subsequence is the right one.
(This is precisely the same recursion pattern we used to solve the subset sum
problem.)

Now for the key question: What do we need to remember about our past
decisions? We can only include A[j] if the resulting subsequence is in increasing
order. If we assume (inductively!) that the numbers previously selected from
A[1.. j — 1] are in increasing order, then we can include A[j] if and only if
A[j ] is larger than the last number selected from A[ 1.. j — 1 ]. Thus, the only
information we need about the past is the last number selected so far. We can
now revise our pictures by erasing everything we don’t need:

6

5

8

9

7

9

3

2

3

8

4

6

2

7

6

8

9

7

9

3

2

3

8

4

6

2

7

So the problem our recursive strategy is actually solving is the following:

Given an integer prev and an array A[1.. n], find the longest increasing

subsequence of A in which every element is larger than prev.

As usual, our recursive strategy requires a base case. Our current strategy
breaks down when we get to the end of the array, because there is no “next
number” to consider. But an empty array has exactly one subsequence, namely,
the empty sequence. Vacuously, every element in the empty sequence is larger
than whatever value you want, and every pair of elements in the empty sequence
appears in increasing order. Thus, the longest increasing subsequence of the
empty array has length 0.

Here’s the resulting recursive algorithm:

85

2. Backtracking

LISBiGGER(prev,Ar 1.. nl):

if n = 0

return 0

else if A[l] < prev

return LISBiGGERfprev, A[2 .. n])

else

skip <— LISBiGGER(prev,A[2.. n])
tafce<— LISbigger(A[ 1],A[2.. n]) + 1
return maxfskip, take}

Okay, but remember that passing arrays around on the call stack is expensive;
let’s try to rephrase everything in terms of array indices, assuming that the array
A[1.. n] is a global variable. The integer prev is typically an array element A[i],
and the remaining array is always a suffix A[j .. n] of the original input array.
So we can reformulate our recursive problem as follows:

Given two indices i and j, where i < j, find the longest increasing
subsequence of A[j.. n] in which every element is larger than A[/].

Let LISbigger{i,j ) denote the length of the longest increasing subsequence of
A[j .. n] in which every element is larger than A[i], Our recursive strategy gives
us the following recurrence:

r o

LISbigger(i,j + 1 )

if j > n
if A[i]>A[j]

LISbigger{i,j) = {

otherwise

Alternatively, if you prefer pseudocode:

LISbigger(i, /):
if j > n

return 0

else if A[l] < prev

return LISbigger(i, ; + 1)

else

skip <— LISbigger(i, j + 1)
take<— LISbiggerQ', j + 1) + 1
return max{skip, take }

Finally, we need to connect our recursive strategy to the original problem:
Finding the longest increasing subsequence of an array with no other constraints.
The simplest approach is to add an artificial sentinel value — oo to the beginning
of the array.

86

27 - Longest Increasing Subsequence, Take 2

LIS(A[ 1.. nl):

m *—oo
return LISbigger(0, 1)

The running time of LISbigger satisfies the Hanoi recurrence T(n) <
2T(n — 1) + 0(1), which as usual implies that T(n) = 0(2 n ). We really shouldn’t
be surprised by this running time; in the worst case, the algorithm examines
each of the 2 n subsequences of the input array.

2.7 Longest Increasing Subsequence, Take 2

This is not the only backtracking strategy we can use to find longest increasing
subsequences. Instead of considering the input array one element at a time,
we could try to construct the output sequence one element at a time. That is,
instead of “Is A[i] is the next element of the output sequence?”, we could ask
directly, “Where is the next element of the output sequence, if any?”

Jumping into the middle of this strategy, we might be faced with the following
picture. Here we just decided to include the 6 just left of the black bar in our
output sequence, and we need to decide which element to the right of the bar
to include next.

31415926 | 535897932384627
Of course, we only need to consider numbers on the right that are bigger than 6.
31415926 | 535897932384627

We have no idea which of those larger numbers is the best choice, and trying to
cleverly figure out the best choice is just going to get us into trouble. Instead,
we enumerate all possibilities by brute force, and continue the decision process
recursively for each one.

87

2. Backtracking

The subset of numbers we can consider as the next element depends only
on the last number we decided to include. Thus, we can simplify our picture of
the decision procedure by discarding all the other numbers that we’ve already
processed.

5 3 5

8 9 7 9

3 2 3

8

4 6 2

7

The remaining numbers are just a suffix of the original input array. Thus, if we
think of the input array A[1.. n] as a global variable, we can formally express
our recursive problem in terms of indices as follows:

Given an index i, find the longest increasing subsequence of A[i .. n ] that
begins with A[i],

Let LISfirst(i) denote the length of the longest increasing subsequence of A[i.. n ]
that begins with A[i], We can now formulate our recursive backtracking strategy
as the following recursive definition:

LISfirst(i ) = 1 + max {LlSfirst(j) | j > i and A[j] >A[i]}

Because we are dealing with sets of natural numbers, we define max0 = 0. Then
we automatically have LISfirst(i ) = 1 if A[j ] > A[t] for all j > i; in particular,
LISfirst(n ) = 1. These are the base cases for our recurrence. We can also express
this recursive definition in pseudocode as follows:

LISfirst(Q:
best <— 0

for j <— i + 1 to n
if A[j] >A[i]

best <— maxjbest, 1 + LISfirst(J)}
return best

Finally, we need to reconnect this recursive algorithm to our original
problem—finding the longest increasing subsequence without knowing its
first element. One natural approach that works is to try all possible first ele¬
ments by brute force. Equivalently, we can add a sentinel element —oo to the
beginning of the array, find the longest increasing subsequence that starts with
the sentinel, and finally ignore the sentinel.

LIS(A[ 1.. nl):

best <— 0
for i <— 1 to n

best <— ma x{best, LISfirst(I)}
return best

LIS(A[ 1.. nl):

A[0] <-oo

return LISfirst(O) — 1

88

2.8. Optimal Binary Search Trees

2.8 Optimal Binary Search Trees

Our final example combines recursive backtracking with the divide-and-conquer
strategy. Recall that the running time for a successful search in a binary search
tree is proportional to the number of ancestors of the target node. 12 As a result,
the worst-case search time is proportional to the depth of the tree. Thus, to
minimize the worst-case search time, the height of the tree should be as small
as possible; by this metric, the ideal tree is perfectly balanced.

In many applications of binary search trees, however, it is more important to
minimize the total cost of several searches rather than the worst-case cost of a
single search. If x is a more frequent search target than y, we can save time by
building a tree where the depth of x is smaller than the depth of y, even if that
means increasing the overall depth of the tree. A perfectly balanced tree is not
the best choice if some items are significantly more popular than others. In fact,
a totally unbalanced tree with depth Q(n) might actually be the best choice!

This situation suggests the following problem. Suppose we are given a
sorted array of keys A[ 1.. n] and an array of corresponding access frequencies
f[l..n]. Our task is to build the binary search tree that minimizes the total
search time, assuming that there will be exactly /[i] searches for each key A[i].

Before we think about how to solve this problem, we should first come
up with a good recursive definition of the function we are trying to optimize!
Suppose we are also given a binary search tree T with n nodes. Let v 1; v 2 ,..., v n
be the nodes of T, indexed in sorted order, so that each node v ; stores the
corresponding key A[t]. Then ignoring constant factors, the total cost of
performing all the binary searches is given by the following expression:

n

Cost(T,f[ 1.. n]) := ^/[t] • #ancestors of v ; in T (*)

i=i

Now suppose v r is the root of T; by definition, v r is an ancestor of every node
in T. If i < r, then all ancestors of v ; except the root are in the left subtree of T.
Similarly, if i > r, then all ancestors of v ; except the root are in the right subtree
of T. Thus, we can partition the cost function into three parts as follows:

Cost(T,f[l..n]) = E/m -

;=i

r—1

^ f[i] ■ #ancestors of v ; in left(T )
i =1

n

+ ^ f[i ] ■ #ancestors of v ; in right(T )
i=r+1

The second and third summations look exactly like our original definition (*)

12 An ancestor of a node v is either the node itself or an ancestor of the parent of v. A proper
ancestor of v is either the parent of v or a proper ancestor of the parent of v.

89

2. Backtracking

for Cost(T,f[l .. n]). Simple substitution now gives us a recurrence for Cost:

n

Cost(T,f[l..n]) = ^f[i] + Cost(left(T),f[l ..r — 1])

1 1 + Cost(right(T),f[r + 1.. n])

The base case for this recurrence is, as usual, n = 0; the cost of performing no
searches in the empty tree is zero.

Now our task is to compute the tree T opt that minimizes this cost function.
Suppose we somehow magically knew that the root of T opt is v r . Then the
recursive definition of Cost(T,f ) immediately implies that the left subtree
le/t(T opt ) must be the optimal search tree for the keys A[1 .. r — 1] and access
frequencies /[I.. r — 1], Similarly, the right subtree right(T opi ) must be the
optimal search tree for the keys A[r + 1.. n] and access frequencies f[r + 1 ..n].
Once we choose the correct key to store at the root, the Recursion Fairy
will construct the rest of the optimal tree.

More generally, let OptCost{i, k) denote the total cost of the optimal search
tree for the interval of frequencies f[i..k]. This function obeys the following
recurrence.

OptCost(i, k )

r

0

k

/ f[i] + min

^ i<r<k

W= !

r OptCost(i,r — 1) )

[ + OptCost(r + l,k)j

if i>k
otherwise

The base case correctly indicates that the minimum possible cost to perform
zero searches into the empty set is zero! Our original problem is to compute
OptCost( 1, n).

This recursive definition can be translated mechanically into a recursive
backtracking algorithm to compute OptCost( 1, n). Not surprisingly, the running
time of this algorithm is exponential. In the next chapter, we’ll see how to
reduce the running time to polynomial, so there’s not much point in computing
the precise running time...

^Analysis

... unless you’re into that sort of thing. Just for the fun of it, let’s figure out
how slow this backtracking algorithm actually is. The running time satisfies the
recurrence

n

nn) = Y t {T(k-l) + nn-k)] + O(n).

k=l

The O(n) term comes from computing the total number of searches /[*]•
Yeah, that’s one ugly recurrence, but we can solve it using exactly the same

go

Exercises

subtraction trick we used before. We replace the 0() notation with an explicit
constant, regroup and collect identical terms, subtract the recurrence for T{n— 1)
to get rid of the summation, and then regroup again.

Tin) =

T{n- 1) =

Tin)-Tin- 1) =
Tin) =

n -1

2 'y~' i T{k) + an

k =o

n—2

2 2>) + a(n-l)

k =0

2T(n — 1) + a
3T(n — 1) + a

Hey, that doesn’t look so bad after all. The recursion tree method immediately
gives us the solution T(n) = 0(3 n ) (or we can just guess and confirm by
induction).

This analysis implies that our recursive algorithm does not examine all
possible binary search trees! The number of binary search trees with n vertices
satisfies the recurrence

72 — 1

Nin) = Y J [Kir-l)-Nin-r)),

r=1

which has the closed-from solution N{n ) = 0(4 n / <Jn). (No, that’s not obvious.)
Our algorithm saves considerable time by searching independently for the optimal
left and right subtrees for each root. A full enumeration of binary search trees
would consider all possible pairs of left and right subtrees; hence the product in
the recurrence for IV(n).

Exercises

i. Describe recursive algorithms for the following generalizations of Subset-
Sum:

(a) Given an array X[1.. n] of positive integers and an integer T, compute
the number of subsets of A whose elements sum to T.

(b) Given two arrays X[l.. n] and W [ 1.. n ] of positive integers and an
integer T, where each W[i] denotes the weight of the corresponding
element X[i], compute the maximum weight subset ofX whose elements
sum to T. If no subset ofX sums to T, your algorithm should return — oo.

2 . Describe recursive algorithms for the following variants of the text segmen¬
tation problem. Assume that you have a subroutine IsWord that takes an

91

2. Backtracking

array of characters as input and returns True if and only if that string is a
“word”.

(a) Given an array A[1.. n] of characters, compute the number of partitions
of A into words. For example, given the string ARTISTOIL, your algorithm
should return 2, for the partitions ARTIST-OIL and ART - IS-TOIL.

(b) Given two arrays A[1.. n] and B[1 ..n] of characters, decide whether A
and B can be partitioned into words at the same indices. For example,
the strings BOTHEARTHANDSATURNSPIN and PINSTARTRAPSANDRAGSLAP
can be partitioned into words at the same indices as follows:

BOT•HEART•HAND•SAT•URNS•PIN
PIN■START■RAPS•AND•RAGS■LAP

(c) Given two arrays A[ 1.. n ] and B [ 1.. n] of characters, compute the number
of different ways that A and B can be partitioned into words at the same
indices.

3 . An addition chain for an integer n is an increasing sequence of integers
that starts with 1 and ends with n, such that each entry after the first
is the sum of two earlier entries. More formally, the integer sequence
x 0 < x 1 < x 2 < ■ ■ ■ < x ( is an addition chain for n if and only if

• *o =

• x ( = n, and

• for every index k > 0 , there are indices i < j <k such that x k = x ; + Xj.

The length of an addition chain is the number of elements minus 1; we
don’t bother to count the first entry. For example, (1,2,3,5,10,20,23,46,
92,184,187,374) is an addition chain for 374 of length 11.

(a) Describe a recursive backtracking algorithm to compute a minimum-
length addition chain for a given positive integer n. Don’t analyze or
optimize your algorithm’s running time, except to satisfy your own
curiosity. A correct algorithm whose running time is exponential in n is
sufficient for full credit. [Hint: This problem is a lot more like n Queens
than text segmentation.]

r (b) Describe a recursive backtracking algorithm to compute a minimum-
length addition chain for a given positive integer n in time that is
sub-exponential in n. [Hint: You may find the results of certain Egyptian
rope-fasteners, Indus-River prosodists, and Russian peasants helpful.]

4 . (a) Let A[1.. m] and B[1.. n] be two arbitrary arrays. A common subsequence

of A and B is both a subsequence of A and a subsequence of B. Give
a simple recursive definition for the function lcs(A,B ), which gives the
length of the longest common subsequence of A and B.

92

Exercises

(b) Let A[l..m] and B[l..n] be two arbitrary arrays. A common super¬
sequence of A and B is another sequence that contains both A and B
as subsequences. Give a simple recursive definition for the function
scs(A,B ), which gives the length of the shortest common supersequence
of A and B.

(c) Call a sequence X[1.. n] of numbers bitonic if there is an index i with
1 < i < n, such that the prefix A [ 1 .. i ] is increasing and the suffix
X[i .. n] is decreasing. Give a simple recursive definition for the function
lbs(A), which gives the length of the longest bitonic subsequence of an
arbitrary array A of integers.

(d) Call a sequence X[l .. n] oscillating if X[i] < X[i + 1] for all even i, and
X[i] > X[i + 1] for all odd i. Give a simple recursive definition for
the function los(A ), which gives the length of the longest oscillating
subsequence of an arbitrary array A of integers.

(e) Give a simple recursive definition for the function sos(A), which gives
the length of the shortest oscillating supersequence of an arbitrary array
A of integers.

Cf) Call a sequence X[1 .. n] convex if 2 • A[i] < X[i — 1] + X[i + 1] for all i.
Give a simple recursive definition for the function lxs(A ), which gives
the length of the longest convex subsequence of an arbitrary array A of
integers.

5 . For each of the following problems, the input consists of two arrays X[1.. k ]
and y[l.. ft] where k < n.

(a) Describe a recursive backtracking algorithm to determine whether X is
a subsequence of Y. For example, the string PPAP is a subsequence of
the string PENPINEAPPLEAPPLEPEN.

(b) Describe a recursive backtracking algorithm to find the smallest number
of symbols that can be removed from Y so that X is no longer a
subsequence. Equivalently, your algorithm should find the longest
subsequence of Y that is not a supersequence of X. For example, after
removing removing two symbols from the string PENPINEAPPLEAPPLEPEN,
the string PPAP is no longer a subsequence.

r (c) Describe a recursive backtracking algorithm to determine whether X
occurs as two disjoint subsequences of Y. For example, the string PPAP ap¬
pears as two disjoint subsequences in the string PENPINEAPPLEAPPLEPEN.

Don’t analyze the running times of your algorithms, except to satisfy your
own curiosity. All three algorithms run in exponential time; we’ll improve
that later, so the precise running time isn’t particularly important.

93

2. Backtracking

6 . This problem asks you to design backtracking algorithms to find the cost of
an optimal binary search tree that satisfies additional balance constraints.
Your input consists of a sorted array A[1.. n] of search keys and an array
/[I.. n] of frequency counts, where f[i] is the number of searches for A[i],
This is exactly the same cost function as described in Section 2 . 8 . But
now your task is to compute an optimal tree that satisfies some additional
constraints.

(a) AVL trees were the earliest self-balancing balanced binary search trees,
first described in 1962 by Georgy Adelson-Velsky and Evgenii Landis. An
AVL tree is a binary search tree where for every node v, the height of
the left subtree of v and the height of the right subtree of v differ by at
most one.

Describe a recursive backtracking algorithm to construct an optimal
AVL tree for a given set of search keys and frequencies.

(b) Symmetric binary B-trees are another self-balancing binary trees, first
described by Rudolf Bayer; these are better known by the name red-
black trees, after a somewhat simpler reformulation by Leo Guibas and
Bob Sedgwick in 1978 . A red-black tree is a binary search tree with the
following additional constraints:

• Every node is either red or black.

• Every red node has a black parent.

• Every root-to-leaf path contains the same number of black nodes.
Describe a recursive backtracking algorithm to construct an optimal
red-black tree for a given set of search keys and frequencies.

(c) AA trees were proposed by proposed by Arne Andersson in 1993 and
slightly simplified (and named) by Mark Allen Weiss in 2000 . AA
trees are also known as left-leaning red-black trees, after a symmetric
reformulation (with different rebalancing algorithms) by Bob Sedgwick
in 2006 . An AA-tree is a red-black tree with one additional constraint:

• No left child is red . 13

Describe a recursive backtracking algorithm to construct an optimal
AA-tree for a given set of search keys and frequencies.

Don’t analyze the running times of your algorithms, except to satisfy your
own curiosity. All three algorithms run in exponential time; we’ll improve
that later, so the precise running time isn’t particularly important.

For more backtracking exercises, see the next chapter!

13 Sedgwick’s reformulation requires that no right child is red. Whatever. Andersson and
Sedgwick are strangely silent on which end of the egg to eat first.

94

Potes enim videre in hac margine, qualiter hoc operati fuimus, scilicet quod
iunximus primum numerum cum secundo, videlicet t cum 2; et secundum cum
tercio; et tercium cum quarto; et quartum cum quinta, et sic deinceps....

[You can see in the margin here how we have worked this; clearly, we combined the
first number with the second, namely t with 2, and the second with the third, and
the third with the fourth, and the fourth with the fifth, and so forth....]

- Leonardo Pisano, Liber Abaci (1202)

Those who cannot remember the past are condemned to repeat it.

- Jorge Agusti'n Nicolas Ruiz de Santayana y Borras,
The Life of Reason, Book I: Introduction and Reason in Common Sense (1905)

You know what a learning experience is?

A learning experience is one of those things that says,

“You know that thing you just did? Don't do that."

— Douglas Adams, The Salmon of Doubt (2002)

3

Dynamic Programming

3.1 Matravrtta

One of the earliest examples of recursion arose in India more than 2000 years ago,
in the study of poetic meter, or prosody. Classical Sanskrit poetry distinguishes
between two types of syllables ( aksara ): light (laghu ) and heavy {guru). In
one class of meters, variously called matravrtta or matrameru or matrachanda,
each line of poetry consists of a fixed number of “beats” {matra), where each
light syllable lasts one beat and each heavy syllable lasts two beats. The formal
study of matra-vrtta dates back to the Chandahsastra, written by the scholar
Pingala between 6oobce and 200 BCE. Pingala observed that there are exactly

five 4-beat meters:-, — • •, • — •, •• —, and • • • •. (Here each “—“

represents a long syllable and each represents a short syllable.) 1

Tn Morse code, a “dah” lasts three times as long as a “dit”, but each “dit” or “dah” is followed
by a pause with the same duration as a “dit”. Thus, each “dit-pause” is a laghu aksara, each

95

3. Dynamic Programming

Although Pingala’s text hints at a systematic rule for counting meters with a
given number of beats , * 2 it took about a millennium for that rule to be stated
explicitly. In the 7 th century ce, another Indian scholar named Virahanka wrote
a commentary on Pirigala’s work, in which he observed that the number of
meters with n beats is the sum of the number of meters with (n — 2 ) beats and
the number of meters with (n — 1) beats. In more modern notation, Virahanka’s
observation implies a recurrence for the total number M(n ) of n-beat meters:

M(n) = M(n - 2) + M(n - 1)

It is not hard to see that M (0) = 1 (there is only one empty meter) and M( 1) = 1
(the only one-beat meter consists of a single short syllable).

The same recurrence reappeared in Europe about 500 years after Virahanka,
in Leonardo Pisano’s 1202 treatise Liber Abaci, one of the most influential
early European works on “algorism”. In full compliance with Stigler’s Law
of Eponymy , 3 the modern Fibonacci numbers are defined using Virahanka’s
recurrence, but with different base cases:

0

if n = 0

1

if n = 1

Fn-1 + F n -2

otherwise

F n+ i for all n.

Backtracking Can Be Slow

The recursive definition of Fibonacci numbers immediately gives us a recur¬
sive algorithm for computing them. Here is the same algorithm written in
pseudocode:

“dah-pause” is a guru aksara, and there are exactly five letters (M, D, R, U, and H) whose codes last
four matra.

2 The Chandahsastra contains two systematic rules for listing all meters with a given number
of syllables, which correspond roughly to writing numbers in binary from left to right (like
Greeks) or from right to left (like Egyptians). The same text includes a recursive algorithm to
compute 2" (the number of meters with n syllables) by repeated squaring, and (arguably) a
recursive algorithm to compute binomial coefficients (the number of meters with k short syllables
and n syllables overall).

3 “No scientific discovery is named after its original discoverer.” In his 1980 paper that gives the
law its name, the statistician Stephen Stigler jokingly claimed that this law was first proposed by
sociologist Robert K. Merton. However, similar statements were previously made by Vladimir
Arnol’d in the 1970’s (“Discoveries are rarely attributed to the correct person.”), Carl Boyer in
1968 (“Clio, the muse of history, often is fickle in attaching names to theorems!”), Alfred North
Whitehead in 1917 (“Everything of importance has been said before by someone who did not
discover it.”), and even Stephen’s father George Stigler in 1966 (“If we should ever encounter a
case where a theory is named for the correct man, it will be noted.”). We will see many other
examples of Stigler’s law in this book.

96

3 .1. Matravrtta

REcFiBo(n):
if n = 0

return 0
else if n = 1
return n

else

return REcFiBo(n — 1) + RecFibo(h — 2)

Unfortunately, this naive recursive algorithm is horribly slow. Except for the
recursive calls, the entire algorithm requires only a constant number of steps:
one comparison and possibly one addition. Let T(n) denote the number of
recursive calls to RecFibo; this function satisfies the recurrence

r(0) = l, r(l) = l, T(n) = r(n —l) + r(n-2) + l,

which looks an awful lot like the recurrence for Fibonacci numbers them¬
selves! Writing out the first several values of T(n) suggests the closed-form
solution T(n) = 2F n+l — 1, which we can verify by induction (hint, hint). So
computing F n using this algorithm takes about twice as long as just counting
to F n . Methods beyond the scope of this book 4 imply that F n = Q(4> n ), where
</> =(i/5 + l)/2 tv 1.61803 is the so-called golden ratio. In short, the running
time of this recursive algorithm is exponential in n.

We can actually see this exponential growth directly as follows. Think of the
recursion tree for RecFibo as a binary tree of additions, with only Os and Is
at the leaves. Since the eventual output is F n , exactly F n of the leaves must
have value 1; these leaves represent the calls to RecRibo(I). An easy inductive
argument (hint, hint) implies that RecFibo(O) is called exactly F n _ ] times. (If
we just want an asymptotic bound, it’s enough to observe that the number
of calls to RecFibo(O) is at most the number of calls to RecFibo(I).) Thus,
the recursion tree has exactly F n + F„_i = F n+1 = 0(F n ) leaves, and therefore,
because it’s a full binary tree, 2 F n+1 — 1 = 0(F n ) nodes altogether.

Memorization: Remember Everything

The obvious reason for the recursive algorithm’s lack of speed is that it com¬
putes the same Fibonacci numbers over and over and over. A single call to
REcFiBo(n) results in one recursive call to REcFiBo(n — 1), two recursive calls
to REcFiBo(n — 2), three recursive calls to RecFibo(h — 3), five recursive calls
to RecFibo(u — 4), and in general F recursive calls to RecFibo(u — k) for
any integer 0 < k < n. Each call is recomputing some Fibonacci number from
scratch.

We can speed up our recursive algorithm considerably just by writing down
the results of our recursive calls and looking them up again if we need them later.

4 See http://algorithms.wtf for notes on solving recurrences.

97

3. Dynamic Programming

Figure 3.1. The recursion tree for computing F 7 , arrows represent recursive calls.

This optimization technique, now known as memoization, is usually credited to
Donald Michie in 1967 , but essentially the same technique was proposed in 1959
by Arthur Samuel . 5

MEMFiBo(n):
if n = 0

return 0
else if n = 1
return 1

else

if F[n ] is undefined

F[n] <— MEMFiBo(n — 1) + MEMFiBo(rt — 2)
return F[n]

Memoization clearly decreases the running time of the algorithm, but by
how much? If we actually trace through the recursive calls made by MemFibo,
we find that the array F[ ] is filled from the bottom up: first F[ 2], then F[3],
and so on, up to F [ n ]. This pattern can be verified by induction: Each entry
F[i] is filled only after its predecessor F[i — 1], If we ignore the time spent in
recursive calls, it requires only constant time to evaluate the recurrence for each
Fibonacci number F ; . But by design, the recurrence for F ; is evaluated only once
for each index i. We conclude that MemFibo performs only O(n) additions, an
exponential improvement over the naive recursive algorithm!

5 Michie proposed that programming languages should support an abstraction he called a
“memo function”, consisting of both a standard function (“rule”) and a dictionary (“rote”), instead
of separately supporting arrays and functions. Whenever a memo function computes a function
value for the first time, it “memorises” (yes, with an R) that value into its dictionary. Michie was
inspired by Samuel’s use of “rote learning” to speed up the recursive evaluation of checkers game
trees; Michie describes his more general proposal as “enabling the programmer to ‘Samuelize’ any
functions he pleases.” (As far as I can tell, Michie never actually used the term “memoisation”.)
Memoization was used even earlier by Claude Shannon’s maze-solving robot “Theseus”, which
he designed and constructed in 1950.

98

3 .1. Matravrtta

Figure 3.2. The recursion tree for F 7 trimmed by memoization. Downward green arrows indicate writing
into the memoization array; upward red arrows indicate reading from the memoization array.

Dynamic Programming: Fill Deliberately

Once we see how the array F[ ] is filled, we can replace the memoized recurrence
with a simple for-loop that intentionally fills the array in order, instead of relying
on a more complicated recursive algorithm to do it for us accidentally.

iTERFiBo(n):

F[ 0]^0
F[ 1]«-1
for i <— 2 to n

F[i]<-F[i-l]+F[i-2]
return F[n]

Now the time analysis is immediate: IterFibo clearly uses O(n) additions and
stores O(n) integers.

This is our first explicit dynamic programming algorithm. The dynamic
programming paradigm was formalized and popularized by Richard Bellman
in the mid- 1950 s, while working at the RAND Corporation, although he was
far from the first to use the technique. In particular, this iterative algorithm
for Fibonacci numbers was already proposed by Virahanka and later Sanskrit
prosodists in the 12 th century, and again by Fibonacci at the turn of the 13 th
century ! 6

6 More general dynamic programming techniques were independently deployed several times
in the late 1930s and early 1940s. For example, Pierre Masse used dynamic programming
algorithms to optimize the operation of hydroelectric dams in France during the Vichy regime.
John von Neumann and Oskar Morgenstern developed dynamic programming algorithms to
determine the winner of any two-player game with perfect information (for example, checkers).
Alan Turing and his cohorts used similar methods as part of their code-breaking efforts at

99

3. Dynamic Programming

Many years after the fact, Bellman claimed that he deliberately chose the
name “dynamic programming” to hide the mathematical character of his work
from his military bosses, who were actively hostile toward anything resembling
mathematical research. * * * 7 The word “programming” does not refer to writing
code, but rather to the older sense of planning or scheduling, typically by filling
in a table. For example, sports programs and theater programs are schedules
of important events (with ads); television programming involves filling each
available time slot with a show (and ads); degree programs are schedules of
classes to be taken (with ads). The Air Force funded Bellman and others to
develop methods for constructing training and logistics schedules, or as they
called them, “programs”. The word “dynamic” was not only a reference to
the multistage, time-varying processes that Bellman and his colleagues were
attempting to optimize, but also a marketing buzzword that would resonate
with the Futuristic Can-Do Zeitgeist™ of post-WWII America. 8 Thanks in part
to Bellman’s proselytizing, dynamic programming is now a standard tool for
multistage planning in economics, robotics, control theory, and several other
disciplines.

Don’t Remember Everything After All

In many dynamic programming algorithms, it is not necessary to retain all
intermediate results through the entire computation. For example, we can
significantly reduce the space requirements of our algorithm IterFibo by
maintaining only the two newest elements of the array:

Bletchley Park. Both Masse’s work and von Neumann and Mergenstern’s work were first published

in 1944, six years before Bellman coined the phrase “dynamic programming”. The details of

Turing’s “Banburismus” were kept secret until the mid-1980s.

7 Charles Erwin Wilson became Secretary of Defense in January 1953, after a dozen years
as the president of General Motors. “Engine Charlie” reorganized the Department of Defense
and significantly decreased its budget in his first year in office, with the explicit goal of running
the Department much more like an industrial corporation. Bellman described Wilson in his 1984
autobiography as follows:

We had a very interesting gentleman in Washington named Wilson. He was secretary of Defense,
and he actually had a pathological fear and hatred of the word "research". I'm not using the term
lightly: I'm using it precisely. His face would suffuse, he would turn red, and he would get violent
if people used the term "research" in his presence. You can imagine how he felt, then, about the
term "mathematical ".... I felt I had to do something to shield Wilson and the Air Force from the fact
that I was really doing mathematics inside the RAND Corporation. What title, what name, could I
choose?

However, Bellman’s first published use of the term “dynamic programming” already appeared in
1952, several months before Wilson took office, so this story is at least slightly embellished.

8 ... and just possibly a riff on the iconic brand name “Dynamic-Tension” for Charles Atlas’s
famous series of exercises, which Charles Roman coined in 1928. Hero of the Beach!

100

*3-2- Aside: Even Faster Fibonacci Numbers

lTERFlBQ2(n):
prev <— 1
curr<— 0
for i <— 1 to n

next <— curr + prev
prev <— curr
curr <— next
return curr

(This algorithm uses the non-standard but consistent base case F_ x = 1 so
that IterFibo2 (0) returns the correct value 0.) Although saving space can be
absolutely crucial in practice, we won’t focus on space issues in this book.

*3.2 Aside: Even Faster Fibonacci Numbers

Although the previous algorithm is simple and attractive, it is not the fastest
algorithm to compute Fibonacci numbers. We can derive a faster algorithm by
exploiting the following matrix reformulation of the Fibonacci recurrence:

'o 1

X

y

1 1

y_

x + y _

In other words, multiplying a two-dimensional vector by the matrix [° {] has
exactly the same effect as one iteration of the inner loop of IterFibo2. It follows
that multiplying by the matrix n times is the same as iterating the loop n times:

O

T— 1

n

i—i

i_

X-r

i

h-i

1

1—1

0

. Fn .

So if we want the nth Fibonacci number, we only need to compute the nth
power of the matrix [° | ]. If we use repeated squaring, 9 computing the nth
power of something requires only 0(log n) multiplications. In this case, that
means 0(log n) 2x2 matrix multiplications, each of which reduces to a constant
number of integer multiplications and additions. Thus, we can compute F n in
only O(logn) integer arithmetic operations.

We can achieve the same speedup using the identity F n = F m F n _ m _ l +
F m+i F n-m> which holds (by induction!) for all integers m and n. In particular,
this identity implies the following mutual recurrence for pairs of adjacent
Fibonacci numbers:

F 2n -i =F 2 i+F 2

An i n—1 n

F 2 n = F n( F n-l + F n+1 ) = F n( 2F n-l + F n )

9 as suggested by Pingala for powers of 2 elsewhere in Chandahsastra

101

3. Dynamic Programming

(We can also derive this mutual recurrence directly from the matrix-squaring
algorithm.) These recurrences translate directly into the following algorithm:

((Compute the pair F n _ v F n ))
FastRecFibo( n):
if n = 1

return 0,1
m <—Ln/2j

hprv, hcur <— FASTREcFiBo(m)

((F m -i, FJ)

prev <— hprv 2 + hcur 2

«f 2m -1»

curr <— hcur • (2 • hprv + hcur )

«f 2m »

next <— prev + curr

((f 2 m+1))

if n is even

return prev, curr

else

return curr, next

Our standard recursion tree technique implies that this algorithm performs only
0 (log n) integer arithmetic operations.

This is an exponential speedup over the standard iterative algorithm, which
was already an exponential speedup over our original recursive algorithm.
Right?

Whoa! Not so fast!

Well, not exactly. Fibonacci numbers grow exponentially fast. The nth Fibonacci
number is approximately n log 10 f w n/ 5 decimal digits long, or n log 2 ^ 2n/3
bits. So we can’t possibly compute F n in logarithmic time — we need Q(n) time
just to write down the answer!

The way out of this apparent paradox is to observe that we can’t perform
arbitrary-precision arithmetic in constant time. Let M(n) denote the time
required to multiply two n-digit numbers. The running time of FastRecFibo
satisfies the recurrence T(n) = T([n/2J) + M(n), which solves to T(n) =
0(M(n)) via recursion trees. The fastest integer multiplication algorithm
known (as of 2018 ) runs in time 0(n log n4 log n ), so that is also the running
time of the fastest algorithm known (as of 2018 ) to compute Fibonacci numbers.

Is this algorithm slower than our “linear-time” iterative algorithms? Actually,
no! Addition isn’t free, either! Adding two n-digit numbers takes O(n) time, so
the running time of the iterative algorithms IterFibo and IterFibo 2 is 0(n 2 ).
(Do you see why?) So FastRecFibo is significantly faster than the iterative
algorithms, just not exponentially faster.

In the original recursive algorithm, the extra cost of arbitrary-precision
arithmetic is overwhelmed by the huge number of recursive calls. The correct

102

3 - 3 - Interpunctio Verborum Redux

recurrence is T(n) = T{n — 1) + T(n — 2) 4- 0{ri), which still has the solution

T(n) = 0 (<n.

3.3 Interpunctio Verborum Redux

For our next dynamic programming algorithm, let’s consider the text segmenta¬
tion problem from the previous chapter. We are given a string A[1.. n] and a
subroutine IsWord that determines whether a given string is a word (whatever
that means), and we want to know whether A can be partitioned into a sequence
of words.

We solved this problem by defining a function Splittable{i ) that returns True
if and only if the suffix A[i .. n] can be partitioned into a sequence of words. We
need to compute Splittable(l). This function satisfies the recurrence

True

if i > n

Splittable{i) = <

n

Y [lsWord(i, j) A Splittable(j + 1))

J =i

otherwise

where IsWord(i,j ) is shorthand for IsWord(A[i .. j]). This recurrence translates
directly into a recursive backtracking algorithm that calls the IsWord subroutine
0 ( 2 n ) times in the worst case.

But for any fixed string A[1.. n], there are only n different ways to call
the recursive function Splittable{i) —one for each value of i between 1 and
n + 1—and only 0(n 2 ) different ways to call IsWord(/, j) —one for each pair
(i, j) such that 1 < i < j < n. Why are we spending exponential time computing
only a polynomial amount of stuff?

Each recursive subproblem is specified by an integer between 1 and n +1, so
we can memoize the function Splittable into an array SplitTable[ 1 ,.n+ 1 ]. Each
subproblem Splittable(i ) depends only on results of subproblems Splittable(j)
where j > i, so the memoized recursive algorithm fills the array in decreasing
index order. If we fill the array in this order deliberately, we obtain the dynamic
programming algorithm shown in Figure 3 . 3 . The algorithm makes 0(n 2 )
calls to IsWord, an exponential improvement over our earlier backtracking
algorithm.

3.4 The Pattern: Smart Recursion

In a nutshell, dynamic programming is recursion without repetition. Dynamic
programming algorithms store the solutions of intermediate subproblems, often
but not always in some kind of array or table. Many algorithms students

103

3. Dynamic Programming

SPLiTTABLE(Ar 1.. nl):

SplitTable[n + 1]«— True
for i <— n down to 1

SplitTable[i ] <— False
for j <— i to n

if IsWord(i, j) and SplitTable[j + 1]

SplitTable[i ] <— True
return SplitTable[ 1]

Figure 3.3. Interpunctio verborum velox

(and instructors, and textbooks) make the mistake of focusing on the table—
because tables are easy and familiar—instead of the much more important
(and difficult) task of finding a correct recurrence. As long as we memoize the
correct recurrence, an explicit table isn’t really necessary, but if the recurrence
is incorrect, we are well and truly hosed.

Dynamic programming is not about filling in tables.

It's about smart recursion!

Dynamic programming algorithms are almost always developed in two distinct
stages.

1 . Formulate the problem recursively. Write down a recursive formula
or algorithm for the whole problem in terms of the answers to smaller
subproblems. This is the hard part. A complete recursive formulation has
two parts:

(a) Specification. Describe the problem that you want to solve recursively,
in coherent and precise English—not how to solve that problem, but
what problem you’re trying to solve. Without this specification, it is
impossible, even in principle, to determine whether your solution is
correct.

(b) Solution. Give a clear recursive formula or algorithm for the whole
problem in terms of the answers to smaller instances of exactly the same
problem.

2 . Build solutions to your recurrence from the bottom up. Write an algo¬
rithm that starts with the base cases of your recurrence and works its way
up to the final solution, by considering intermediate subproblems in the
correct order. This stage can be broken down into several smaller, relatively
mechanical steps:

(a) Identify the subproblems. What are all the different ways your re¬
cursive algorithm can call itself, starting with some initial input? For
example, the argument to RecFibo is always an integer between 0 and n.

104

3 - 5 - Warning: Greed is Stupid

(b) Choose a memoization data structure. Find a data structure that can
store the solution to every subproblem you identified in step (a). This is
usually but not always a multidimensional array.

(c) Identify dependencies. Except for the base cases, every subproblem
depends on other subproblems—which ones? Draw a picture of your
data structure, pick a generic element, and draw arrows from each of
the other elements it depends on. Then formalize your picture.

(d) Find a good evaluation order. Order the subproblems so that each one
comes after the subproblems it depends on. You should consider the
base cases first, then the subproblems that depends only on base cases,
and so on, eventually building up to the original top-level problem. The
dependencies you identified in the previous step define a partial order
over the subproblems; you need to find a linear extension of that partial
order. Be careful!

(e) Analyze space and running time. The number of distinct subproblems
determines the space complexity of your memoized algorithm. To
compute the total running time, add up the running times of all possible
subproblems, assuming deeper recursive calls are already memoized. You
can actually do this immediately after step (a).

(f) Write down the algorithm. You know what order to consider the
subproblems, and you know how to solve each subproblem. So do that!
If your data structure is an array, this usually means writing a few nested
for-loops around your original recurrence, and replacing the recursive
calls with array look-ups.

Of course, you have to prove that each of these steps is correct. If your recurrence
is wrong, or if you try to build up answers in the wrong order, your algorithm
won’t work!

3.5 Warning: Greed is Stupid

If we’re incredibly lucky, we can bypass all the recurrences and tables and so forth,
and solve the problem using a greedy algorithm. Like a backtracking algorithm, a
greedy algorithm constructs a solution through a series of decisions, but it makes
those decisions directly, without solving at any recursive subproblems. While this
approach seems very natural, it almost never works; optimization problems that
can be solved correctly by a greedy algorithm are quite rare. Nevertheless, for
many problems that should be solved by backtracking or dynamic programming,
many students’ first intuition is to apply a greedy strategy.

For example, a greedy algorithm for the longest increasing subsequence
problem might look for the smallest element of the input array, accept that

105

3. Dynamic Programming

element as the start of the target subsequence, and then recursively look for the
longest increasing subsequence to the right of that element. If this sounds like
a stupid hack to you, pat yourself on the back. It isn’t even close to a correct
solution.

Everyone should tattoo the following sentence on the back of their hands,
right under all the rules about logarithms and big-Oh notation:

Greedy algorithms never work!

Use dynamic programming instead!

What, never?

No, never!

What, never ?

Well... hardly ever. 10

Because the greedy approach is so incredibly tempting, but so rarely correct,
I strongly advocate the following policy in any algorithms course, even (or
perhaps especially') for courses that do not normally ask for proofs of correctness. 12

You will not receive any credit for any greedy algorithm,
on any homework or exam, even if the algorithm is correct,
without a formal proof of correctness.

Moreover, the vast majority of problems for which students are tempted to
submit a greedy algorithm are actually best solved using dynamic programming.
So I always offer the following advice to my algorithms students.

Whenever you write—or even think —the word “greeDY”,
your subconscious is telling you to use DYnamic programming.

Even for problems that can be correctly solved by greedy algorithms, it’s usually
more productive to develop a backtracking or dynamic programming algorithm
first. First make it work; then make it fast. We will see techniques for proving
greedy algorithms correct in the next chapter.

“Greedy methods hardly ever work! Then give three cheers, and one cheer more, for the
careful Captain of the Pinafore ! Then give three cheers, and one cheer more, for the Captain of
the Pinafore ! u

U I am a very good theoretical computer scientist; specifically, a geometric algorithm specialist.
“Introducing this policy in my own algorithms courses significantly improved students’ grades,
because it significantly reduced the number of times they submitted incorrect greedy algorithms.

106

3.6. Longest Increasing Subsequence

3.6 Longest Increasing Subsequence

Another problem we considered in the previous chapter was computing the
length of the longest increasing subsequence of a given arrayA [1 .. n] of numbers.
We developed two different recursive backtracking algorithms for this problem.
Both algorithms run in 0(2 n ) time in the worst case; both algorithms can be
sped up significantly via dynamic programming.

First Recurrence: Is This Next?

Our first backtracking algorithm evaluated the function LISbigger{i,j), which
we defined as the length of the longest increasing subsequence of A[j .. n] in
which every element is larger than A[i]. We derived the following recurrence
for this function:

r

0

LISbigger(i, j) = <

LISbigger(i,j + 1)

( LISbigger(i,j + 1) )
max \ >

1 1 + LISbigger(j, j + l)j

if j > n
if A[i]>A[j]

otherwise

To solve the original problem, we can add a sentinel value A[0] = —00 to the
array and compute LISbigger( 0,1).

Each recursive subproblem is identified by two indices i and j, so there are
only 0(n 2 ) distinct recursive subproblems to consider. We can memoize the re¬
sults of these subproblems into a two-dimensional array LISbigger[0.. n, 1 .. n]. 13
Moreover, each subproblem can be solved in 0(1) time, not counting recursive
calls, so we should expect the final dynamic programming algorithm to run in
0(n 2 ) time.

The order in which the memoized recursive algorithm fills this array is
not immediately clear; all we can tell from the recurrence is that each entry
LISbigger[i, j] is filled in after the entries LlSbigger[i,j + l] and LISbigger[j, j +1 ]
in the next column, as indicated on the left in Figure 3 . 4 .

Fortunately, this partial information is enough to give us a valid, evaluation
order. If we fill the table one column at a time, from right to left, then whenever
we reach an entry in the table, the entries it depends on are already available.
This may not be the order that the recursive algorithm would use, but it works,
so we’ll go with it. The right figure in Figure 3.4 illustrates this evaluation order,
with a double arrow indicating the outer loop and single arrows indicating the

13 In fact, we only need half of this array, because we always have i < j. But even if we cared
about constant factors in this class (we don’t), this would be the wrong time to worry about them.
First make it work; then make it better.

107

3. Dynamic Programming

Figure 3.4. Subproblem dependencies for longest increasing subsequence, and a valid evaluation order

inner loop. In this case, the single arrows are bidirectional, because the order
that we use to fill each column doesn’t matter.

And we’re done! Pseudocode for our dynamic programming algorithm is
shown below; as expected, our algorithm clearly runs in 0(n 2 ) time. If necessary,
we can reduce the space bound from 0(n 2 ) to O(rt) by maintaining only the
two most recent columns of the table, LISbigger[-, j] and LISbigger[-, j + I ]. 14

LIS(A[ 1.. n])?

A[0] * - 00 {{Add a sentinel))

for i «— 0 to n {{Base cases))

LISbigger[i, n + 1]«— 0
for j <— n down to 1

for i <— 0 to j — 1 ((... or whatever))

keep <— 1 + LISbigger[j, j + 1]
skip <— LISbigger[i, j + 1]
if A[i]>A[j]

LISbigger[i,j ] <— skip

else

LISbigger[i, j ] <— max{keep,skip}
return LISbigger[ 0,1]

Second Recurrence: What’s Next?

Our second backtracking algorithm evaluated the function LISfirst{i ), which
we defined as the length of the longest increasing subsequence of A[i .. n] that
begins with A[i], We derived the following recurrence for this function:

LISfirst{i ) = 1 + max {LISfirst{j) | j > i and A[j] >A[i]}

Here, we assume that max0 = 0, so that the base cases like LISfirst{n) = 1 fall
out of the recurrence automatically. To solve the original problem, we can add
a sentinel value A[0] = —00 to the array and compute LISfirst{0 ) — 1.

In this case, recursive subproblems are indicated by a single index t, so we
can memoize the recurrence into a one-dimensional array LISfirst[ 1.. n ]. Each

14 See, I told you not to worry about constant factors yet!

108

37- Edit Distance

entry LISfirst[i] depends only on entries LISfirst[j ] with j > i, so we can fill
the array in decreasing index order. To compute each LISfirst[i ], we need to
consider LISfirst[j ] for all indices j > i, but we don’t need to consider those
indices j in any particular order. The resulting dynamic programming algorithm
runs in 0(n 2 ) time and uses O(n) space.

LIS 2 (Ari..nl):

A[ 0 ] = —00

((Add a sentinel))

for i <— n downto 0
LISfirst[i ] <— 1
for j <— i + 1 to n

((... or whatever))

if A[j] > A[i] and 1 + LISfirst[j] > LISfirst[i ]

LISfirst[i ]

<— 1 + LISfirst[j ]

return LISfirst[0 ] — 1

((Don't count the sentinel))

3.7 Edit Distance

The edit distance between two strings is the minimum number of letter inser¬
tions, letter deletions, and letter substitutions required to transform one string
into the other. For example, the edit distance between FOOD and MONEY is at
most four:

FOOD MOOD -> MOND MONED -> MONEY

This distance function was independently proposed by Vladimir Levenshtein in
1965 (working on coding theory), Taras Vintsyuk in 1968 (working on speech
recognition), and Stanislaw Ulam in 1972 (working with biological sequences).
For this reason, edit distance is sometimes called Levenshtein distance or Ulam
distance (but strangely, never “Vintsyuk distance”).

We can visualize this editing process by aligning the strings one above the
other, with a gap in the first word for each insertion and a gap in the second
word for each deletion. Columns with two different characters correspond to
substitutions. In this representation, the number of editing steps is just the
number of columns that do not contain the same character twice.

F 0 0 D
MONEY

It’s fairly obvious that we can’t transform FOOD into MONEY in three steps, so
the edit distance between FOOD and MONEY is exactly four. Unfortunately, it’s not
so easy in general to tell when a sequence of edits is as short as possible. For
example, the following alignment shows that the distance between the strings
ALGORITFIM and ALTRUISTIC is at most 6 . Is that the best we can do?

109

3. Dynamic Programming

ALGOR I THM
AL TRUISTIC

Recursive Structure

To develop a dynamic programming algorithm to compute edit distance, we
first need to develop a recurrence. Our alignment representation for edit
sequences has a crucial “optimal substructure” property. Suppose we have the
gap representation for the shortest edit sequence for two strings. If we remove
the last column, the remaining columns must represent the shortest edit
sequence for the remaining prefixes. We can easily prove this observation
by contradiction: If the prefixes had a shorter edit sequence, gluing the last
column back on would gives us a shorter edit sequence for the original strings.
So once we figure out what should happen in the last column, the Recursion
Fairy can figure out the rest of the optimal gap representation.

Said differently, the alignment we are looking for represents a sequence of
editing operations, ordered (for no particular reason) from right to left. Solving
the edit distance problem requires making a sequence of decisions, one for each
column in the output alignment. In the middle of this sequence of decisions, we
have already aligned a suffix of one string with a suffix of the other.

ALGOR

I

T

H

M

ALTRU

I

S

T

I

C

Because the cost of an alignment is just the number of mismatched columns,
our remaining decisions don’t depend on the editing operations we’ve already
chosen; the only depend on the prefixes we haven’t aligned yet.

ALGOR

ALTRU

Thus, for any two input strings A[1 ..m] and B[1.. n], we can formulate the edit
distance problem recursively as follows: For any indices i and j, let Edit(i,j )
denote the edit distance between the prefixes A[1 .. i] and B[1We need to
compute Edit(jn, n).

Recurrence

When i and j are both positive, there are exactly three possibilities for the last
column in the optimal alignment of A[1.. i] and B[1

• Insertion: The last entry in the bottom row is empty. In this case, the edit
distance is equal to Edit{i — !,_/) + 1. The +1 is the cost of the final insertion,

no

37- Edit Distance

and the recursive expression gives the minimum cost for the remaining
alignment.

ALGO

R

ALTRU

• Deletion: The last entry in the top row is empty. In this case, the edit
distance is equal to Edit(i, j — 1) + 1. The +1 is the cost of the final deletion,
and the recursive expression gives the minimum cost for the remaining
alignment.

ALGOR

ALTR

U

• Substitution: Both rows have characters in the last column. If these two
characters are different, then the edit distance is equal to Edit(i — 1,; —1) +1.
If these two characters are equal, the substitution is free, so the edit distance
is Edit{i — 1 ,; — 1).

ALGO

R

ALGO

R

ALTR

U

ALT

R

This generic case analysis breaks down if either i = 0 or j = 0, but those
boundary cases are easy to handle directly.

• Transforming the empty string into a string of length j requires j insertions,
so Edit(0,j ) = j.

• Transforming a string of length i into the empty string requires i deletions,
so Edit(i, 0) = i.

As a sanity check, both of these base cases correctly indicate that the edit
distance between the empty string and the empty string is zero!

We conclude that the Edit function satisfies the following recurrence:

(.
i

Edit(i,j)= -

J

mm <

, Edit{i

Edit(i — 1,;) + 1
Edit{i,j — 1) + 1

-lJ-l) + [A[i]#B[j]L

if; = 0
if i = 0

otherwise

Dynamic Programming

Now that we have a recurrence, we can transform it into a dynamic programming
algorithm following our usual mechanical recipe.

• Subproblems: Each recursive subproblem is identified by two indices
0 < i < m and 0 < j < n.

in

3. Dynamic Programming

• Memoization structure: So we can memoize all possible values of Edit(i,j)
in a two-dimensional array Edit[ 0.. m, 0 .. n].

• Dependencies: Each entry Edit[i,j] depends only on its three neighboring
entries Edit[i — 1 ,j], Edit[i,j — 1], and Edit[i — l,j — 1],

• Evaluation order: If we fill this array in standard row-major order—row by
row from top down, each row from left to right—then whenever we reach an
entry in the array, all the entries it depends on are already available. (This
isn’t the only evaluation order we could use, but it works, so let’s go with it.)

• Space and time: The memoization structure uses O(mn) space. We can
compute each entry Edit[i,j] in 0(1) time once we know its predecessors,
so the overall algorithm runs in 0(mn) time.

Here is the resulting dynamic programming algorithm:

EditDistance(A( 1.. ml,.BT1.. nj):

for j <— 0 to n
Edit[0,j] <— j

for i «— 1 to m
Edit[i, 0] <— i
for j <— 1 to n

ins <— Edit[i — 1, j] + 1
del <— Edit[i,j — 1] + 1
if A[i] = B[j]

rep <— Edit[i — 1, j — 1]

else

rep <— Edit[i — 1, j — 1] + 1
Edit[i,j ] <— min {ins, del, rep}
return Edit[m, n ]

This algorithm is most commonly attributed to Robert Wagner and Michael
Fischer, who described the algorithm in 1974 . However, in full compliance
with Stigler’s Law of Eponymy, either identical or more general algorithms
were independently discovered by Taras Vintsyuk in 1968 , V. M. Velichko and
N. G. Zagoruyko in 1970 , David Sankoff in 1972 , Peter Sellers in 1974 , and

112

37- Edit Distance

almost certainly several others. 15 Interestingly, none of these authors cite either
Levenshtein or Ulam!

The memoization table for the input strings ALGORITHM and ALTRUISTIC is
shown below. Bold numbers indicate places where characters in the two strings
are equal. The edit distance between ALGORITHM and ALTRUISTIC is indeed six!

A

L

G

0

R

I

T H

M

0

1

2

3

4

5

6

7 8

9

i\

A

1

0 1

2

3

-4

->5

6 7

8

1

L

2

1

0

->1

2

3

4 5 6

7

1

* \ \ \

T

3

2

1

1

2

3

4

4 5

6

1

i

i\ \

R

4

3

2

2

2

2

3

-4 5

6

1

1

s, \, \

U

5

4

3

3

3

3

3

4 5

6

i

s,l N

u\ \ \ \

I

6

5

4

4

4

4

3

4^5

6

1\ \ \

S

7

6

5

5

5

5

4

4 5

6

J,

1\ \ \

T

8

7

6

6

6

6

5

4 5

6

a\i

I

9

8

7

7

7

7

6

5 5

6

1

•1 ' s \vl s ''\ >1

|\J\

C

10

9

8

8

8

8

7

6 6

6

The arrows in this table indicate which predecessor(s) actually define
each entry. Each direction of arrow corresponds to a different edit operation:
horizontal=deletion, vertical=insertion, and diagonal=substitution. Bold red
diagonal arrows indicate “free” substitutions of a letter for itself. Any path of
arrows from the top left corner to the bottom right corner of this table represents
an optimal edit sequence between the two strings. The example memoization
array contains exactly three directed paths from the top left corner to the bottom
right corner, each indicating a different sequence of six edits transforming
ALGORITHM into ALTRUISTIC, as shown on the next page.

15 This algorithm is sometimes also incorrectly attributed to Saul Needleman and Christian
Wunsch in 1970. “The Needleman-Wunsch algorithm” more commonly refers to the standard
dynamic programming algorithm for computing the longest common subsequence of two strings
(or equivalently, the edit distance where only insertions and deletions are permitted) in 0 (mn )
time, but that attribution is also incorrect! In fact, Needleman and Wunsch’s algorithm computes
(weighted) longest common subsequences (possibly with gap costs) in 0 {m 2 n 2 ) time, using a
different recurrence. Sankoff explicitly describes his 0 (mn)-time algorithm as an improvement
of Needleman and Wunsch’s algorithm.

113

3. Dynamic Programming

ALGORI THM
ALTRUISTIC

ALGOR I THM
AL TRUISTIC

ALGOR I THM
ALT RUISTIC

Our EditDistance algorithm does not actually compute or store any arrows
in the table, but the arrow (s) leading into any entry in the table can be
reconstructed on the fly in 0(1) time from the numerical values. Thus, once
we’ve filled in the table, we can reconstruct the shortest edit sequence in 0(n+m)
additional time.

3.8 Subset Sum

Recall that the Subset Sum problem asks, whether any subset of a given array
X[1.. n] of positive integers sums to a given integer T. In the previous chapter,
we developed a recursive Subset Sum algorithm that can be reformulated as
follows. Fix the original input array X[ 1.. n] and the original target sum T, and
define the boolean function

SS(i, t) = True if and only if some subset of X[i .. n] sums to t.

This function satisfies the following recurrence:

r

if t = 0

if t < 0 or i > n

True

SS(i, t)= < False

SS(i + l, t) V SS(i + l,t —X[i]) otherwise

We can transform this recurrence into a dynamic programming algorithm
following the usual boilerplate.

• Subproblems: Each subproblem is described by an integer i such that
1 < i < n + 1, and an integer t <T. However, subproblems with t < 0 are
trivial, so it seems rather silly to memoize them. 16 Indeed, we can modify
the recurrence so that those subproblems never arise:

True

if t = 0
if i > n
if t>X[i]

SS{i + l,t) V SS(i + l,t-X[i])

otherwise

l6 Yes, I’m breaking my own rule against premature optimization.

114

3-9- Optimal Binary Search Trees

• Data structure: We can memoize our recurrence into a two-dimensional
array S[1.. n + 1,0.. T], where S[i, t] stores the value of SS(i, t ).

• Evaluation order: Each entry S[i, t] depends on at most two other entries,
both of the form SS[i + 1, •]. So we can fill the array by considering rows
from bottom to top in the outer loop, and considering the elements in each
row in arbitrary order in the inner loop.

• Space and time: The memoization structure uses O(nT) space. If S[i + 1, t]
and S[i + 1, t—X[t]] are already known, we can compute S[i, t] in constant
time, so the algorithm runs in O(nT) time.

Here is the resulting dynamic programming algorithm:

SubsetSum(X f 1 .. n], T):

S[n + 1,0] <— True

for t <— 1 to T

S[n + 1, t]«— False

for i «— n downto 1
S[i,0] = True
for t <— 1 to X[i] — 1

S[i, t]«— S[i + 1, t] ((Avoid the case t < 0))
for t «—X[i] to T

return S[l, T ]

The worst-case running time O(nT) for this algorithm is a significant
improvement over the 0(2 n )-time recursive backtracking algorithm when T is
small. 17 However, if the target sum T is significantly larger than 2 n , this iterative
algorithm is actually slower than the naive recursive algorithm, because it’s
wasting time solving subproblems that the recursive algorithm never considers.
Dynamic programming isn’t always an improvement! 18

3.9 Optimal Binary Search Trees

The final problem we considered in the previous chapter was the optimal binary
search tree problem. The input is a sorted array A[1.. n] of search keys and an
array/[I.. n] of frequency counts, where f[i] is the number of times we will

17 Even though the Subset Sum problem is NP-complete, this time bound does not imply that
P=NP, because T is not necessary bounded by a polynomial function of the input size.

l8 In the 1967 research memorandumf!) where he proposed memo functions, Donald Michie
wrote, “To tabulate values of a function which will not be needed is a waste of space, and to
recompute the same values more than once is a waste of time.” But in fact, tabulating values of a
function that will not be needed is also a waste of time\

115

3. Dynamic Programming

search for A[i]. Our task is to construct a binary search tree for that set such
that the total cost of all the searches is as small as possible.

Fix the frequency array /, and let OptCost(i, k) denote the total search time
in the optimal search tree for the subarray A[i.. k]. We derived the following
recurrence for the function OptCosf.

r

0

if i > fc

OptCost(i, k)

£/m

+ min

i<r<k

(OptCost(i, r — 1)

| + OptCost(r + 1, k )

otherwise

You can probably guess what we’re going to do with this recurrence eventually,
but let’s rid of that ugly summation first.

For any pair of indices i < k, let F(i, k) denote the total frequency count for
all the keys in the interval A[i ..k]:

k

;=i

This function satisfies the following simple recurrence:

F{i,k)

|/[i] if i = fc

[f(i, k — 1) + f[k] otherwise

We can compute all possible values of F(i,k) in 0(n 2 ) time using—you guessed
it!—dynamic programming! The usual mechanical steps give us the following
dynamic programming algorithm:

InitF( f |~ 1 ..nl):

for i <— 1 to n

F[i, i — 1] <— 0
for k <— i to n

F[i,k] *— F[i,k—1]+ f[k]

We will use this short algorithm as an initialization subroutine. This initialization
allows us to simplify the original OptCost recurrence as follows:

0

OptCost(i, k)

-

F[i, k] + min

i<r<k

j OptCost(i, r — 1)

[ + OptCost(r + 1, k )

if f > fc
otherwise

Now let’s turn the crank.

3-9- Optimal Binary Search Trees

• Subproblems: Each recursive subproblem is specified by two integers i
and k, such that 1 < i < n + 1 and 0 < k < n.

• Memoization: We can store all possible values of OptCost in a two-
dimensional array OptCost[ 1.. n + 1,0.. n]. (Only the entries OptCost[i, j]
with j > i — 1 will actually be used, but whatever.)

• Dependencies: Each entry OptCost[i, k] depends on the entries OptCost[i,
j — 1] and OptCost[j + 1, k\, for all j such that i <j<k. In other words,
each table entry depends on all entries either directly to the left or directly
below.

The following subroutine fills the entry OptCost[i, k], assuming all the
entries it depends on have already been computed.

ComputeOptCost(i, k ):

OptCost[i, fc]<— oo
for r «— i to k

tmp <— OptCost[i, r — 1] + OptCost[r + 1, k ]
if OptCost[i, fc] > tmp
OptCost[i,k ] <— tmp
OptCost[i, fc] <— OptCost[i, fc] + F[i, fc]

• Evaluation order: There are at least three different orders that can be
used to fill the array. The first one that occurs to most students is to scan
through the table one diagonal at a time, starting with the trivial base cases
OptCost[i, i — 1] and working toward the final answer OptCo.st[l, n], like so:

OPTiMALBSTff Tl ..nl):

InitF(/[ 1 .. n])
for i <— 1 to n + 1

OptCost[i, i— 1] <— 0
for d <— 0 to n — 1

for i <— 1 to n — d. ((... or whatever))

ComputeOptCost(i, i + d)
return OptCost[l, n]

We could also traverse the array row by row from the bottom up, traversing
each row from left to right, or column by column from left to right, traversing
each columns from the bottom up.

117

3. Dynamic Programming

OptimalBST 2 ( f Tl.. n]):

OPTiMALBSTsffri ..nT):

InitF(/[ 1 ..n])

InitF(/[ 1 ..n])

for i <— n + 1 downto 1

for j <— 0 to n + 1

OptCost[i,i — 1] <— 0

OptCost[j + 1, j] <— 0

for j «— i to n

for i <— j downto 1

ComputeOptCost(i, j)

ComputeOptCost(i, j)

return OptCost[l,n]

return OptCost[l, n]

As before, we can illustrate these evaluation orders using a double-lined
arrow to indicate the outer loop and single-lined arrows to indicate the inner
loop. The bidirectional arrows in the first evaluation order indicate that the
order of the inner loops doesn’t matter.

• Time and space: The memoization structure uses 0(n 2 ) space. No matter
which evaluation order we choose, we need O(n) time to compute each
entry OptCost[i,k], so our overall algorithm runs in 0(n 3 ) time.

As usual, we could have predicted the final space and time bounds directly from
the original recurrence:

0

if i>k

OptCost(i, k) = <

V

F[i, k] + min

i<r<k

j OptCost(i, r — 1)

| + OptCost(r + 1, k)

otherwise

The OptCost function has two arguments, each of which can take on roughly n
different values, so we probably need a data structure of size 0(n 2 ). On the
other hand, there are three variables in the body of the recurrence (i, k, and r),
each of which can take roughly n different values, so it should take 0 (n 3 ) time
to compute everything.

3.10 Dynamic Programming on Trees

So far, all of our dynamic programming examples use multidimensional arrays
to store the results of recursive subproblems. However, as the next example
shows, this is not always the most appropriate date structure to use.

An independent set in a graph is a subset of the vertices with no edges
between them. Finding the largest independent set in an arbitrary graph is
extremely hard; in fact, this is one of the canonical NP-hard problems we will

3 . 10 . Dynamic Programming on Trees

study in Chapter 12 . But in some special classes of graphs, we can find largest
independent sets quickly. In particular, when the input graph is a tree with n
vertices, we can actually compute the largest independent set in 0(n) time.

Suppose we are given a tree T. Without loss of generality, suppose T is a
rooted tree; that is, there is a special node in T called the root, and all edges are
implicitly directed away from this vertex. (If T is an unrooted tree—a connected
acyclic undirected graph—we can choose an arbitrary vertex as the root.) We
call vertex w a descendant of vertex v if the unique path from w to the root
includes v; equivalently, the descendants of v are v itself and the descendants
of the children of v. The subtree rooted at v consists of all the descendants of v
and the edges between them.

For any node v in T, letMIS(v) denote the size of the largest independent set
in the subtree rooted at v. Any independent set in this subtree that excludes v
itself is the union of independent sets in the subtrees rooted at the children of v.
On the other hand, any independent set that includes v necessarily excludes all
of v’s children, and therefore includes independent sets in the subtrees rooted
at v’s grandchildren. Thus, the function MIS obeys the following recurrence,
where the nonstandard notation w [v means “ w is a child of v”:

MIS(v) = max i ^M/S(v), 1 + M7S(x)

( w|v w|v x|w

We need to compute MIS{r ), where r is the root of T.

Figure 3.5. Computing the maximum independent set in a tree

What data structure should we use to memoize this recurrence? The most
natural choice is the tree T itself! Specifically, for each vertex v in T, we store
the result of MIS(v) in a new field v.MIS. (In principle, we could use an array
instead, but then we’d have to pointers back and forth between each node and
its corresponding array entry, so why bother?)

What’s a good order to consider the subproblems? The subproblem associ¬
ated with any node v depends on the subproblems associated with the children
and grandchildren of v. So we can visit the nodes in any order we like, provided
that every vertex is visited before its parent; in particular, we can use a standard
post-order traversal.

What’s the running time of the algorithm? The non-recursive time associated
with each node v is proportional to the number of children and grandchildren

3. Dynamic Programming

of v; this number can be very different from one vertex to the next. But we can
turn the analysis around: Each vertex contributes a constant amount of time to
its parent and its grandparent! Because each vertex has at most one parent and
at most one grandparent, the algorithm runs in O(n) time.

Here is the resulting dynamic programming algorithm. Yes, it’s still recursive,
because that’s the most natural way to implement a post-order tree traversal.

MIS(v):
skipv <— 0

for each child w of v

skipv <— skipv + MIS(w)
keepv <— 1

for each grandchild x of v
keepv <— keepv + x.MIS
v.MIS <— ma x{keepv,skipv}
return v.MIS

We can derive an even simpler linear-time algorithm by defining two separate
functions over the nodes of T:

• Let MISyes{v ) denote the size of the largest independent set of the subtree
rooted at v that includes v.

• Let MISno{v) denote the size of the largest independent set of the subtree
rooted at v that excludes v.

Again, we need to compute ma x{MISyes(r),MISno(r)}, where r is the root of T.
The first two functions satisfy the following mutual recurrence:

MISyes{v) = 1 + ^ ~' i MISno(w )

w[v

MISno(y ) = ^ max {MISyes(w),MISno(w)}

w[v

Again, we can memoize these functions into the tree itself, by defining two
new fields for each vertex. A straightforward post-order traversal evaluates
both functions at every node in O(n) time. The following algorithm not only
memoizes both function values at v, it also returns the larger of those two
values.

MIS(v):

v.MISno <— 0
v.MISyes <— 1
for each child w of v

v.MISno <— v.MISno + MIS(w)
v.MISyes <— v.MISyes + w.MISno
return ma x{v.MISyes,v.MISno}

120

Exercises

In the second line of the inner loop, we are using the value w.MISno that was
memoized by the recursive call in the previous line.

Exercises

For all of the following exercises—and more generally when developing any
new dynamic programming algorithm—I strongly recommend following the
steps outlined in Section 3 . 4 . In particular, don’t even start thinking about
tables or for-loops until you have a complete recursive solution, including a clear
English specification of the recursive subproblems you are actually solving . 19
First make it work; then make it fast.

Sequences/Arrays

3 . In a previous life, you worked as a cashier in the lost Antarctican colony of
Nadira, spending the better part of your day giving change to your customers.
Because paper is a very rare and valuable resource in Antarctica, cashiers
were required by law to use the fewest bills possible whenever they gave
change. Thanks to the numerological predilections of one of its founders,
the currency of Nadira, called Dream Dollars, was available in the following
denominations: $ 1 , $ 4 , $ 7 , $ 13 , $ 28 , $ 52 , $ 91 , $ 36 s. * 2 °

*(a) The greedy change algorithm repeatedly takes the largest bill that does
not exceed the target amount. For example, to make $122 using the
greedy algorithm, we first take a $91 bill, then a $28 bill, and finally
three $1 bills. Give an example where this greedy algorithm uses more
Dream Dollar bills than the minimum possible. [Hint: It may be easier
to write a small program than to work this out by hand.]

(b) Describe and analyze a recursive algorithm that computes, given an
integer k, the minimum number of bills needed to make k Dream
Dollars. (Don’t worry about making your algorithm fast; just make sure
it’s correct.)

(c) Describe a dynamic programming algorithm that computes, given an
integer k, the minimum number of bills needed to make k Dream Dollars.
(This one needs to be fast.)

19 In my algorithms classes, any dynamic programming solution that does not include an
English specification of the underlying recursive subproblems automatically gets a score of zero,
even if the solution is otherwise perfect. Introducing this policy in my own algorithms courses
significantly improved students’ grades, because it significantly reduced the number of times they
submitted incorrect (or incoherent) dynamic programming algorithms.

2 °For more details on the history and culture of Nadira, including images of the various
denominations of Dream Dollars, see http://moneyart.biz/dd/.

121

3. Dynamic Programming

4 . Describe efficient algorithms for the following variants of the text segmen¬
tation problem. Assume that you have a subroutine IsWord that takes
an array of characters as input and returns True if and only if that string
is a “word”. Analyze your algorithms by bounding the number of calls to
IsWORD.

(a) Given an array A[1.. n] of characters, compute the number of partitions
of A into words. For example, given the string ARTISTOIL, your algorithm
should return 2, for the partitions ARTIST-OIL and ART - IS-TOIL.

(b) Given two arrays A[1.. n ] and B[1 ..n] of characters, decide whether A
and B can be partitioned into words at the same indices. For example,
the strings BOTHEARTHANDSATURNSPIN and PINSTARTRAPSANDRAGSLAP
can be partitioned into words at the same indices as follows:

BOT•HEART•HAND•SAT•URNS•PIN
PIN■START■RAPS•AND•RAGS•LAP

(c) Given two arrays A[ 1.. n ] and B [ 1.. n ] of characters, compute the number
of different ways that A and B can be partitioned into words at the same
indices.

5 . Suppose you are given an array A[1.. n] of numbers, which may be positive,
negative, or zero, and which are not necessarily integers.

(a) Describe and analyze an algorithm that finds the largest sum of elements
in a contiguous subarray A[i

(b) Describe and analyze an algorithm that finds the largest product of
elements in a contiguous subarray A[i.._/].

For example, given the array [— 6 ,12, —7,0,14, —7,5] as input, your first
algorithm should return 19, and your second algorithm should return 504.

sum=19

-6

12

-7

0

14

-7

5

- S' -

product=504

Given the one-element array [—374] as input, your first algorithm should
return 0, and your second algorithm should return 1. (The empty interval is
still an interval!) For the sake of analysis, assume that comparing, adding,
or multiplying any pair of numbers takes 0(1) time.

[Hint: Part (a) has been a standard computer science interview question
since at least the mid-1980s. You can find many correct solutions on the
web; the problem even has its own Wikipedia page! But at least in 2016, a
significant fraction of the solutions I found on the web for part (b) were
either slower than necessary or actually incorrect.]

122

Exercises

6 . This exercise explores variants of the maximum-subarray problem (Prob¬
lem 5 ). In all cases, your input consists of an array A[1 ..n] of real numbers
(which could be positive, negative, or zero) and possibly an additional
integer X > 0.

(a) Wrapping around: Suppose A is a circular array In this setting, a
“contiguous subarray” can be either an interval A[i .. j ] or a suffix followed
by a prefix A[i .. n ] • A[ 1 ..j]. Describe and analyze an algorithm that
finds a contiguous subarray of A with the largest sum.

(b) Long subarrays only: Describe and analyze an algorithm that finds a
contiguous subarray of A of length at least X that has the largest sum.
(Assume X < n.)

(c) Short subarrays only: Describe and analyze an algorithm that finds a
contiguous subarray of A of length at mostX that has the largest sum.

(d) The Price Is Right: Describe and analyze an algorithm that finds a
contiguous subarray of A with the largest sum less than or equal to X.

(e) Describe a faster algorithm for Problem 6 (d) when every number in the
array A is non-negative.

7 . This exercise asks you to develop efficient algorithms to find optimal subse¬
quences of various kinds. A subsequence is anything obtained from a sequence
by extracting a subset of elements, but keeping them in the same order; the
elements of the subsequence need not be contiguous in the original sequence.
For example, the strings C, DAMN, YAI0AI, and DYNAMICPROGRAMMING are all
subsequences of the string DYNAMICPROGRAMMING.

[Hint: Exactly one of these problems can be solved in O(n) time using a
greedy algorithm.]

(a) Let A[l..m] and B[1 .. n] be two arbitrary arrays. A common sub¬
sequence of A and B is another sequence that is a subsequence of both
A and B . Describe an efficient algorithm to compute the length of the
longest common subsequence of A and B.

(b) Let A[1.. m] and B[1 .. n] be two arbitrary arrays. A common super¬
sequence of A and B is another sequence that contains both A and B as
subsequences. Describe an efficient algorithm to compute the length of
the shortest common supersequence of A and B.

(c) Call a sequence X[1.. n] of numbers bitonic if there is an index i with
1 < i < n, such that the prefix X[1.. /] is increasing and the suffix
X[i .. n] is decreasing. Describe an efficient algorithm to compute the
length of the longest bitonic subsequence of an arbitrary array A of
integers.

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3. Dynamic Programming

(d) Call a sequence X[1 ..n] of numbers oscillating if X[i] < X[i + 1] for
all even i, and X[t] > X[i + 1] for all odd i. Describe an efficient
algorithm to compute the length of the longest oscillating subsequence
of an arbitrary array A of integers.

(e) Describe an efficient algorithm to compute the length of the shortest
oscillating supersequence of an arbitrary array A of integers.

Cf) Call a sequenceX[l.. n] of numbers convex if 2-X[i] <X[i—1]+X[£ + 1]
for all i. Describe an efficient algorithm to compute the length of the
longest convex subsequence of an arbitrary array A of integers.

(g) Call a sequence X[1.. n] of numbers weakly increasing if each element
is larger than the average of the two previous elements; that is, 2-X[i] >
X[i — 1] +X[£ — 2] for all i > 2. Describe an efficient algorithm to
compute the length of the longest weakly increasing subsequence of an
arbitrary array A of integers.

(h) Call a sequence X[1.. n] of numbers double-increasing ifX[i] > X[i—2]
for all i > 2. (In other words, a double-increasing sequence is a perfect
shuffle of two increasing sequences.) Describe an efficient algorithm to
compute the length of the longest double-increasing subsequence of an
arbitrary array A of integers.

(i) Recall that a sequence X[1.. n ] of numbers is increasing if X[i] <X[i + l]
for all i. Describe an efficient algorithm to compute the length of the
longest common increasing subsequence of two given arrays of integers. For
example, (1,4,5,6, 7,9) is the longest common increasing subsequence
of the sequences (3,1,4,1,5,9,2,6,5,3,5,8,9, 7,9,3) and (1,4,1,4,2,
1 ,3,5,6,2,3,7,3,0,9,5).

8 . A shuffle of two strings X and Y is formed by interspersing the characters

into a new string, keeping the characters of X and Y in the same order.

For example, the string BANANAANANAS is a shuffle of the strings BANANA and

ANANAS in several different ways.

banana ANAN as ban ANA ana NAS b AN an A a NA na S

Similarly, the strings PRODGYRNAMAMMIINCG and DYPRONGARMAMMICING are

both shuffles of DYNAMIC and PROGRAMMING:

proDgYrNAMammiJnCg dy PRO n G a R m AMM ic ING

(a) Given three strings A[1.. m], B[1.. n], and C[1.. m + n], describe and
analyze an algorithm to determine whether C is a shuffle of A and B.

(b) A smooth shuffle of X and Y is a shuffle of X and Y that never uses
more than two consecutive symbols of either string. For example,

124

Exercises

• ^R^Oy 1 NARAmMMj ; cNG j s a smooth shuffle of the strings DYNAMIC and
PROGRAMMING.

. DY PR N OGR A A M MM IC ING i s a shuffle of DYNAMIC and PROGRAMMING, but
it is not a smooth shuffle (because of the substrings OGR and ING).

• XX X X X X XX XX XX XX X X X XX is a smooth shuffle of the strings XXXXXXX
and XXXXXXXXXXX.

• There is no smooth shuffle of the strings XXXX and XXXXXXXXXXXX.

Describe and analyze an algorithm to decide, given three strings X, Y,
and Z, whether Z is a smooth shuffle of X and Y.

9 . For each of the following problems, the input consists of two arrays X[1 .. k]

and 7[1.. n] where k <n.

(a) Describe and analyze an algorithm to decide whether X is a subsequence
of Y. For example, the string PPAP is a subsequence of the string
PENPINEAPPLEAPPLEPEN.

(b) Describe and analyze an algorithm to find the smallest number of symbols
that can be removed from Y so that X is no longer a subsequence.
Equivalently, your algorithm should find the longest subsequence of Y
that is not a supersequence of X. For example, after removing removing
two symbols from the string PENPINEAPPLEAPPLEPEN, the string PPAP is
no longer a subsequence.

r (c) Describe and analyze an algorithm to determine whether X occurs as
two disjoint subsequences of Y. For example, the string PPAP appears as
two disjoint subsequences in the string PENPINEAPPLEAPPLEPEN.

(d) Suppose the input also includes a third array C[1.. n] of numbers, which
maybe positive, negative, or zero, where C[i] is the cost of Y[i]. Describe
and analyze an algorithm to compute the minimum-cost occurrence of
X as a subsequence of Y. That is, we want to find an array I[1 ..k] such
that I[j] < I[j + 1 ] andX[I[j]] = Y[j] for every index j, and the total
cost 2 j=i C[j] is as small as possible.

(e) Describe and analyze an algorithm to compute the total number of
(possibly overlapping) occurrences of X as a subsequence of Y. For
purposes of analysis, assume that we can add two arbitrary integers in
0(1) time. For example, the string PPAP appears exactly 23 times as a
subsequence of the string PENPINEAPPLEAPPLEPEN. If all characters inX
and Y are equal, your algorithm should return

(f) What is the running time of your algorithm for part (d) if adding two
€ -bit integers requires 0 (f) time?

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3. Dynamic Programming

10 . Describe and analyze an efficient algorithm to find the length of the longest
contiguous substring that appears both forward and backward in an input
string T[l..n]. The forward and backward substrings must not overlap.
Here are several examples:

• Given the input string ALGORITHM, your algorithm should return 0.

• Given the input string RECURSION, your algorithm should return 1, for
the substring R.

• Given the input string R EDI V IDE, your algorithm should return 3, for the
substring EDI. (The forward and backward substrings must not overlap!)

• Given the input string DYNAMICPROGRAMMINGMANYTIMES, your algorithm
should return 4, for the substring YNAM. (In particular, it should not
return 6 , for the subsequence YNAM I R).

n. A palindrome is any string that is exactly the same as its reversal, like I, or
DEED, or RACECAR, or AMANAPLANACATACANALPANAMA.

(a) Describe and analyze an algorithm to find the length of the longest
subsequence of a given string that is also a palindrome.

For example, the longest palindrome subsequence of the string
MAHDYNAMICPROGRAMZLETMESHOWYOUTHEM is MHYMRORMYHM; thus, given
that string as input, your algorithm should return 11 .

(b) Describe and analyze an algorithm to find the length of the shortest
supersequence of a given string that is also a palindrome. For example,
the shortest palindrome supersequence of TWENTYONE is TWENT 0 Y 0 TNEWT,
so given the string TWENTYONE as input, your algorithm should return 13.

(c) Any string can be decomposed into a sequence of palindromes. For
example, the string BUBBASEESABANANA (“Bubba sees a banana.”) can be
broken into palindromes in the following ways (and 65 others):

BUB • BASEESAB • ANANA
B • U • BB • ASEESA • B • ANANA
BUB • B • A • SEES • ABA • N • ANA
B • U • BB • A • S • EE • S • A • B • A • NAN • A
B'U'B'B'A'S'E'E'S'A^B^A^N'A^N^A

Describe and analyze an efficient algorithm to find the smallest number
of palindromes that make up a given input string. For example, given
the input string BUBBASEESABANANA, your algorithm should return 3.

(d) Describe and analyze an efficient algorithm to find the largest integer k
such that a given string can be split into palindromes of length at least k.
For example

126

Exercises

• Given the string PALINDROME, your algorithm should return 1.

• Given the string BUBBASEESABANANA, your algorithm should return 3,
for the partition BUB • BASEESAB • ANANA.

• Given a string of n identical symbols, your algorithm should return n.

(e) Describe and analyze an efficient algorithm to find the number of
different ways that a given string can be decomposed into palindromes.
For example:

• Given the string PALINDROME, your algorithm should return 1.

• Given the string BUBBASEESABANANA, your algorithm should return
70.

• Given a string of n identical symbols, your algorithm should return
2 n ~ 1 .

(f) A metapalindrome is a decomposition of a string into a sequence of
palindromes, such that the sequence of palindrome lengths is itself a
palindrome. For example:

BOB • S • MAM • ASEESA • UKU • L • ELE

is a metapalindrome for the string BOBSMAMASEESAUKULELE, whose length
sequence is the palindrome (3,1,3,6,3,1,3). Describe and analyze an
efficient algorithm to find the length of the shortest metapalindrome for a
given string. For example, given the input string BOBSMAMASEESAUKULELE,
your algorithm should return n.

12 . Suppose you are given an array A[1.. n] of positive integers. An increas¬
ing back-and-forth subsequence is an sequence of indices I[l ..£] with the
following properties:

• 1 < I[j] < n for all j.

• A[/[;']] <A[I[j + 1]] for all j < £.

• If 7[j] is even, then I[j + 1] > /[_/].

• If I[j] is odd, then I[j + 1] < /[;'].

Less formally, suppose we are given an array of n squares, each containing a
positive integer. Suppose we place a token on one of the squares, and then
repeatedly move the token left (if it’s on an odd-indexed square) or right
(if it’s on an even-indexed square), always moving from a smaller number
to a larger number. Then the sequence of token positions is an increasing
back-and-forth subsequence.

Describe an algorithm to compute the length of the longest increasing
back-and-forth subsequence of a given array of n red and blue integers. For
example, given the input array

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3. Dynamic Programming

1

1

8

5

6

3

6

4

4

8

3

9

1

2

2

3

9

4

0

1 2 3 4 5 6 7 8 g 10 11 12 13 14 15 16 17 18 19 20

your algorithm should return the integer 9, which is the length of the
following increasing back-and-forth subsequence:

0

1

2

3

4

6

7

8

9

20 1 15 18 10 6 4 3 13

13 . Suppose we want to typeset a paragraph of text onto a piece of paper (or if
you insist, a computer screen). The text consists of a sequence of n words,
where the ith word has length l [t]. We want to break the paragraph into
several lines of total length exactly L. For example, according to TpX, the
program used to typeset these notes, the paragraph you are reading right
now is approximately 11.94794 cm & 4.7055 inches wide.

Depending on how the paragraph is broken into lines of text, we must
insert different amounts of white space between the words. The paragraph
should be fully justified, meaning that the first character on each line starts
at the left margin, and except for the last line, the last character on each line
ends at the right margin. There must be at least one unit of white space
between any two words on the same line. See the paragraph you are reading
right now ? Just like that.

Define the slop of a paragraph layout as the sum over all lines, except
the last, of the cube of the amount of extra white-space in each line, not
counting the one unit of required space between each adjacent pair of words.
Specifically, if a line contains words i through j, then the slop of that line is
defined to be (l — j + i — XijU,- • Describe a dynamic programming
algorithm to print the paragraph with minimum slop.

14 . You and your eight-year-old nephew Elmo decide to play a simple card
game. At the beginning of the game, the cards are dealt face up in a long
row. Each card is worth a different number of points. After all the cards are
dealt, you and Elmo take turns removing either the leftmost or rightmost
card from the row, until all the cards are gone. At each turn, you can decide
which of the two cards to take. The winner of the game is the player that
has collected the most points when the game ends.

Having never taken an algorithms class, Elmo follows the obvious greedy
strategy—when it’s his turn, Elmo always takes the card with the higher
point value. Your task is to find a strategy that will beat Elmo whenever
possible. (It might seem mean to beat up on a little kid like this, but Elmo
absolutely hates it when grown-ups let him win.)

(a) Prove that you should not also use the greedy strategy. That is, show
that there is a game that you can win, but only if you do not follow the
same greedy strategy as Elmo.

128

Exercises

(b) Describe and analyze an algorithm to determine, given the initial se¬
quence of cards, the maximum number of points that you can collect
playing against Elmo.

*(c) When Elmo was four, he used an even simpler strategy—on his turn,
he always chose his next card uniformly at random. That is, if there
was more than one card left on his turn, he would take the leftmost
card with probability 1 / 2 , and the rightmost card with probability 1 / 2 .
Describe an algorithm to determine, given the initial sequence of cards,
the maximum expected number of points you can collect playing against
four-year-old-Elmo.

(d) Five years later, thirteen-year-old Elmo has become a much stronger
player. Describe and analyze an algorithm to determine, given the initial
sequence of cards, the maximum number of points that you can collect
playing against a perfect opponent.

15 . It’s almost time to show off your flippin’ sweet dancing skills! Tomorrow is
the big dance contest you’ve been training for your entire life, except for that
summer you spent with your uncle in Alaska hunting wolverines. You’ve
obtained an advance copy of the list of n songs that the judges will play
during the contest, in chronological order. Yessssssssss!

You know all the songs, all the judges, and your own dancing ability
extremely well. For each integer k, you know that if you dance to the kth
song on the schedule, you will be awarded exactly Score[k ] points, but then
you will be physically unable to dance for the next Wait[k ] songs (that is,
you cannot dance to songs k + 1 through k + Wait[k]). The dancer with the
highest total score at the end of the night wins the contest, so you want your
total score to be as high as possible.

Describe and analyze an efficient algorithm to compute the maximum
total score you can achieve. The input to your sweet algorithm is the pair of
arrays Score[ 1.. n] and Wait[ 1..n],

16 . The new swap-puzzle game Candy Swap Saga XIII involves n cute animals
numbered from 1 to n. Each animal holds one of three types of candy:
circus peanuts. Heath bars, and Cioccolateria Gardini chocolate truffles. You
also have a candy in your hand; at the start of the game, you have a circus
peanut.

To earn points, you visit each of the animals in order from 1 to n. For
each animal, you can either keep the candy in your hand or exchange it
with the candy the animal is holding.

• If you swap your candy for another candy of the same type, you earn
one point.

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3. Dynamic Programming

• If you swap your candy for a candy of a different type, you lose one point.
(Yes, your score can be negative.)

• If you visit an animal and decide not to swap candy, your score does not
change.

You must visit the animals in order, and once you visit an animal, you can
never visit it again.

Describe and analyze an efficient algorithm to compute your maximum
possible score. Your input is an array C[1 ..n], where C[i] is the type of
candy that the ith animal is holding.

17 . Lenny Rutenbar, the founding dean of the new Maksymilian R. Levchin
College of Computer Science, has commissioned a series of snow ramps on
the south slope of the Orchard Downs sledding hill 21 and challenged Bill
Kudeki, head of the Department of Electrical and Computer Engineering, to
a sledding contest. Bill and Lenny will both sled down the hill, each trying to
maximize their air time. The winner gets to expand their department/college
into both Siebel Center and the new ECE Building; the loser has to move
their entire department/college in the Boneyard culvert next to Loomis Lab.

Whenever Lenny or Bill reaches a ramp while on the ground, they can
either use that ramp to jump through the air, possibly flying over one or
more ramps, or sled past that ramp and stay on the ground. Obviously, if
someone flies over a ramp, they cannot use that ramp to extend their jump.

(a) Suppose you are given a pair of arrays Ramp[ 1.. n] and Length[ 1.. n],
where Ramp[i ] is the distance from the top of the hill to the ith ramp,
and Length[i ] is the distance that any sledder who takes the ith ramp will
travel through the air. Describe and analyze an algorithm to determine
the maximum total distance that Lenny or Bill can spend in the air.

(b) The university lawyers heard about Lenny and Bill’s little bet and
immediately objected. To protect the university from either lawsuits
or sky-rocketing insurance rates, they impose an upper bound on the
number of jumps that either sledder can take. Describe and analyze
an algorithm to determine the maximum total distance that Lenny or
Bill can spend in the air with at most k jumps, given the original arrays
Ramp[ 1.. n] and Length[ 1.. n] and the integer k as input.

*(c) When the lawyers realized that imposing their restriction didn’t immedi¬
ately shut down the contest, they added a new restriction: No ramp can
be used more than once! Disgusted by the legal interference, Lenny and
Bill give up on their bet and decide to cooperate to put on a good show

“The north slope is faster, but too short for an interesting contest.

130

Exercises

for the spectators. Describe and analyze an algorithm to determine the
maximum total distance that Lenny and Bill can spend in the air, each
taking at most k jumps (so at most 2k jumps total), and with each ramp
used at most once.

18 . Farmers Boggis, Bunce, and Bean have set up an obstacle course for Mr. Fox.
The course consists of a long row of booths, each with a number painted on
the front with bright red paint. Formally, Mr. Fox is given an array A[1.. n],
where A[i] is the number painted on the front of the ith booth. Each number
A[i] could be positive, negative, or zero. Everyone agrees with the following
rules:

• At each booth, Mr. Fox must say either “Ring!” or “Ding!”.

• If Mr. Fox says “Ring!” at the ith booth, he earns a reward of A[i]
chickens. (If A[i] < 0, Mr. Fox pays a penalty of —A[ i ] chickens.)

• If Mr. Fox says “Ding!” at the ith booth, he pays a penalty of A[i] chickens.
(If A[i] < 0, Mr. Fox earns a reward of —A[i] chickens.)

• Mr. Fox is forbidden to say the same word more than three times in a
row. For example, if he says “Ring!” at booths 6 , 7 , and 8 , then he must
say “Ding!” at booth 9 .

• All accounts will be settled at the end, after Mr. Fox visits every booth
and the umpire calls “Hot box!” Mr. Fox does not actually have to carry
chickens through the obstacle course.

• Finally, if Mr. Fox violates any of the rules, or if he ends the obstacle
course owing the farmers chickens, the farmers will shoot him.

Describe and analyze an algorithm to compute, the largest number of
chickens that Mr. Fox can earn by running the obstacle course, given the
array A[ 1.. n] of numbers as input. [Hint: Watch out for the burning pine
cone!]

19 . Dance Dance Revolution is a dance video game, first introduced in Japan
by Konami in 1998 . Players stand on a platform marked with four arrows,
pointing forward, back, left, and right, arranged in a cross pattern. During
play, the game plays a song and scrolls a sequence of n arrows (<-, f, < 1 ,
or -*) from the bottom to the top of the screen. At the precise moment
each arrow reaches the top of the screen, the player must step on the
corresponding arrow on the dance platform. (The arrows are timed so that
you’ll step with the beat of the song.)

You are playing a variant of this game called “Vogue Vogue Revolution”,
where the goal is to play perfectly but move as little as possible. When an
arrow reaches the top of the screen, if one of your feet is already on the

131

3. Dynamic Programming

correct arrow, you are awarded one style point for maintaining your current
pose. If neither foot is on the right arrow, you must move one (and only one)
foot from its current location to the correct arrow on the platform. If you
ever step on the wrong arrow, or fail to step on the correct arrow, or move
more than one foot at a time, or move either foot when you are already
standing on the correct arrow, all your style points are taken away and you
lose the game.

How should you move your feet to maximize your total number of style
points? For purposes of this problem, assume you always start with your left
foot on 4 - and your right foot on - 4 , and that you’ve memorized the entire
sequence of arrows. For example, if the sequence istt^i<-4<-4, you
can earn 5 style points by moving your feet as shown below:

t

t

L

L

R |

R]

4

l

Begin!

t

Style point!

Style point!

Style point! Style point! Style point!

(a) Prove that for any sequence of n arrows, it is possible to earn at least
n/4 — 1 style points.

(b) Describe an efficient algorithm to find the maximum number of style
points you can earn during a given WR routine. The input to your
algorithm is an array Arrow[l.. n] containing the sequence of arrows.

20 . Consider the following solitaire form of Scrabble. We begin with a fixed,
finite sequence of tiles; each tile has both a letter and a numerical value. At
the start of the game, we draw the first seven tiles from the sequence and
put them into our hand. In each turn, we form an English word from some
or all of the tiles in our hand, place those tiles on the table, and receive the
total value of those tiles as points. (If no English word can be formed from
the tiles in our hand, the game immediately ends.) Then we repeatedly draw
the next tile from the start of the sequence until either (a) we have seven
tiles in our hand, or (b) the sequence is empty. (Sorry, no double/triple
word/letter scores, bingos, blanks, or passing.) Our goal is to obtain as
many points as possible.

For example, consider the following sequence of 20 tiles:

Given this sequence of tiles at the beginning of the game, we can earn 68
points as follows:

• We initially draw [h] [nT] [x7] [aT] [nT] [aT] [pT|.

• Play the word [n7] [aT] [g] [aT] [pT] f° r 9 points, leaving [x7] in hand.

132

Exercises

• Draw the next five tiles | u 5 11 d 3 11 1 2 11 d 3 11 k 8 | .

• Play the word [1)7] [17] [17] []d 7] for 15 points, leaving [1<7] [x7| in hand.

• Draw the next five tiles | u 5 11 b 4 11 l 2 11 ^ 11 k 8 | .

• Play the word | b 4 11 u 5 11 l 2 11 k 8 | for 19 points, leaving [1<7] | x 8 11 A 1 | in hand.

• Draw the last three tiles [17] [7] | n 2 | .

• Play the word | A a 11 n 2 11 k 8 11 h 5 | for 16 points, leaving [1<7] [aT| in our hand.

• Play the word [~a7] [x 7] for 9 points, emptying our hand and ending the
game.

(a) Suppose the sequence of tiles is represented by two arrays Letter[l.. n ],
containing a sequence of letters between A and Z, and Value[ A.. Z], where
Value[l ] is the value of any tile with letter l. Design and analyze an
efficient algorithm to compute the maximum number of points that can
be earned from the given sequence of tiles.

(b) Now suppose two tiles with the same letter might have different values.
Now the tile sequence is represented by two arrays Letter[l.. n] and
Value[l .. n], where Value[i ] is the value of the ith tile. Design and
analyze an efficient algorithm to compute the maximum number of
points that can be earned from the given sequence of tiles.

In both problems, the output is a single number: the maximum possible
score. Assume that you can find all English words that can be made from
any set of at most seven tiles, along with the point values of those words, in
0 ( 1 ) time.

21 . Suppose we are given a set L of n line segments in the plane, where each
segment has one endpoint on the line y = 0 and one endpoint on the line
y = 1 , and all 2 n endpoints are distinct.

(a) Describe and analyze an algorithm to compute the largest subset of L in
which no pair of segments intersects.

(b) Describe and analyze an algorithm to compute the largest subset of L in
which every pair of segments intersects.

Now suppose we are given a set L of n line segments in the plane, where
both endpoints of each segment lie on the unit circle x 2 + y 2 = 1 , and all
2n endpoints are distinct.

(c) Describe and analyze an algorithm to compute the largest subset of L in
which no pair of segments intersects.

(d) Describe and analyze an algorithm to compute the largest subset of L in
which every pair of segments intersects.

133

3. Dynamic Programming

22 . Let P be a set of n points evenly distributed on the unit circle, and let S
be a set of m line segments with endpoints in P. The endpoints of the m
segments are not necessarily distinct; n could be significantly smaller than
2m.

(a) Describe an algorithm to find the size of the largest subset of segments
in S such that every pair is disjoint. Two segments are disjoint if they do
not intersect even at their endpoints.

(b) Describe an algorithm to find the size of the largest subset of segments
in S such that every pair is interior-disjoint. Two segments are interior-
disjoint if their intersection is either empty or an endpoint of both
segments.

(c) Describe an algorithm to find the size of the largest subset of segments
in S such that every pair intersects.

(d) Describe an algorithm to find the size of the largest subset of segments
in S such that every pair crosses. Two segments cross if they intersect
but not at their endpoints.

For full credit, all four algorithms should run in O(mn') time.

23 . You are driving a bus along a highway, full of rowdy, hyper, thirsty students
and a soda fountain machine. Each minute that a student is on your bus,
that student drinks one ounce of soda. Your goal is to drop the students
off quickly, so that the total amount of soda consumed by all students is as
small as possible.

You know how many students will get off of the bus at each exit. Your
bus begins somewhere along the highway (probably not at either end) and
move s at a constant speed of 37.4 miles per hour. You must drive the
bus along the highway; however, you may drive forward to one exit then
backward to an exit in the opposite direction, switching as often as you like.
(You can stop the bus, drop off students, and turn around instantaneously.)

Describe an efficient algorithm to drop the students off so that they drink
as little soda as possible. Your input consists of the bus route (a list of the
exits, together with the travel time between successive exits), the number of
students you will drop off at each exit, and the current location of your bus
(which you may assume is an exit).

24 . Let’s define a summary of two strings A and B to be a concatenation of
substrings of the following form:

• ASNA indicates a substring SNA of only the first string A.

• ♦FOO indicates a common substring F00 of both strings.

• TBAR indicates a substring BAR of only the second string B.

134

Exercises

A summary is valid if we can recover the original strings A and B by
concatenating the appropriate substrings of the summary in order and
discarding the delimiters ▲, ♦, and ▼. Each regular character has length 1,
and each delimiter ▲, ♦, or ▼ has some fixed non-negative length A. The
length of a summary is the sum of the lengths of its symbols.

For example, each of the following strings is a valid summary of the
strings KITTEN and KNITTING:

• ♦KTNflTTAETI^NTG has length 9+ 7A.

• ♦KTNfITTAENTING has length 10 +5A.

• ♦KAITTENYNITTING has length 13 + 3A.

• AKITTENYKNITTING has length 14 + 2A.

Describe and analyze an algorithm that computes the length of the
shortest summary of two given strings A[1 .. m] and B[1.. n]. The delimiter
length A is also part of the input to your algorithm. For example:

• Given strings KITTEN and KNITTING and A = 0, your algorithm should
return 9.

• Given strings KITTEN and KNITTING and A = 1, your algorithm should
return 15.

• Given strings KITTEN and KNITTING and A = 2, your algorithm should
return 18.

25 . Vankin’s Mile is an American solitaire game played on an n x n square grid.
The player starts by placing a token on any square of the grid. Then on
each turn, the player moves the token either one square to the right or one
square down. The game ends when player moves the token off the edge of
the board. Each square of the grid has a numerical value, which could be
positive, negative, or zero. The player starts with a score of zero; whenever
the token lands on a square, the player adds its value to his score. The
object of the game is to score as many points as possible.

For example, given the grid below, the player can score 8 —6+7—3 + 4 =
10 points by placing the initial token on the 8 in the second row, and then
moving down, down, right, down, down. (This is not the best possible score
for this grid of numbers.)

-1

7

-8

10

-5

-4

-9

8

11

-6

0

5

-2

-6

11

-6

7

-7

4

V'

7 =

^3

II

-3

7

1

-6

4

MM

-9

135

3. Dynamic Programming

(a) Describe and analyze an efficient algorithm to compute the maximum
possible score for a game of Vankin’s Mile, given the nxn array of values
as input.

(b) In the European version of this game, appropriately called Vankin’s
Kilometer, the player can move the token either one square down, one
square right, or one square left in each turn. However, to prevent infinite
scores, the token cannot land on the same square more than once.
Describe and analyze an efficient algorithm to compute the maximum
possible score for a game of Vankin’s Kilometer, given the n x n array of
values as input. 22

26 . Suppose you are given an m x n bitmap as an array M[ 1.. n, 1.. n] of Os
and Is. A solid block in M is a subarray of the form M[i .. i', j .. /] in which
all bits are equal. A solid block is square if it has the same number of rows
and columns.

(a) Describe an algorithm to find the maximum area of a solid square block
in M in 0(n 2 ) time.

(b) Describe an algorithm to find the maximum area of a solid block in M
in 0(n 3 ) time.

(c) Describe an algorithm to find the maximum area of a solid block in M
in 0(n 2 logn) time. [Hint: Divide and conquer.]

r (d) Describe an algorithm to find the maximum area of a solid block in M
in 0(n 2 ) time.

27 . Suppose you are given an array M[1.. n, 1.. n] of numbers, which may be
positive, negative, or zero, and which are not necessarily integers. Describe
an algorithm to find the largest sum of elements in any rectangular subarray
of the form M[i .. i',j ../]. For full credit, your algorithm should run in
0(n 3 ) time. [Hint: See problem 5.]

28 . Describe and analyze an algorithm that finds the maximum-area rectangular
pattern that appears more than once in a given bitmap. Specifically, given
a two-dimensional array M[l..n, l..n] of bits as input, your algorithm
should output the area of the largest repeated rectangular pattern in M.
For example, given the bitmap shown on the left in the figure below, your
algorithm should return the integer 195, which is the area of the 15x13
doggo. (Although it doesn’t happen in this example, the two copies of the
repeated pattern might overlap.)

22 If we also allowed upward movement, the resulting game (Vankin’s Fathom?) would be

NP-hard.

136

Exercises

ijs" "

7 17i

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iiS*

™r-;™

■■ ■■■"

ry

■■■■■■■■■

11 \::%s

“Tl

s , ■■

fail

p\

j-s!,™?

y ■■ j ij

■■■■■

2H

(a) For full credit, describe an algorithm that runs in 0(n 5 ) time.
r (b) For extra credit, describe an algorithm that runs in 0(n 4 ) time.

(c) For extra extra credit, describe an algorithm that runs in 0(n 3 polylog n)
time.

29 . Let P be a set of points in the plane in convex position. Intuitively, if a
rubber band were wrapped around the points, then every point would touch
the rubber band. More formally, for any point p in P, there is a line that
separates p from the other points in P. Moreover, suppose the points are
indexed P[l], P[2],..., P[n] in counterclockwise order around the “rubber
band”, starting with the leftmost point P[l],

This problem asks you to solve a special case of the traveling salesman
problem, where the salesman must visit every point in P, and the cost of
moving from one point p e P to another point q e P is the Euclidean distance

HI-

(a) Describe a simple algorithm to compute the shortest cyclic tour of P.

(b) A simple tour is one that never crosses itself. Prove that the shortest tour
of P must be simple.

(c) Describe and analyze an efficient algorithm to compute the shortest tour
of P that starts at the leftmost point P[l] and ends at the rightmost
point P[r],

(d) Describe and analyze an efficient algorithm to compute the shortest tour
of P, with no restrictions on the endpoints.

* 30 . Describe and analyze an algorithm to solve the traveling salesman problem
in 0(2" poly(n)) time. Given an undirected n-vertex graph G with weighted
edges, your algorithm should return the weight of the lightest cycle in G
that visits every vertex exactly once, or 00 if G has no such cycles. [Hint:
The obvious recursive backtracking algorithm takes 0(n!) time.]

31 . Let W = {w 1 ,w 2 , ■ ■., w n } be a finite set of strings over some fixed alphabet S.
An edit center for W is a string C € E* such that the maximum edit distance

137

3. Dynamic Programming

from C to any string in W is as small as possible. The edit radius of W is the
maximum edit distance from an edit center to a string in W. A set of strings
may have several edit centers, but its edit radius is unique.

EditRadiusCW ) := min max Edit(w,C)

CeE* weW

EditCenter{W) := argmin max Edit(w, C)

CeE* weW

(a) Describe and analyze an efficient algorithm to compute the edit radius
of three given strings.

**(b) Describe and analyze an efficient algorithm to approximate the edit
radius of an arbitrary set of strings within a factor of 2. (Computing the
exact edit radius is NP-hard unless the number of strings is fixed.)

* 32 . Let D[1.. n] be an array of digits, each an integer between 0 and 9. A digital
subsequence of D is a sequence of positive integers composed in the usual
way from disjoint substrings of D. For example, the sequence 3,4,5,6, 8 ,
9,32,38,46,64,83,279 is a digital subsequence of the first several digits
of n:

3 , 1, 4 , 1, 5 , 9 , 2 , 6 , 5 , 3 , 5 , 8 , 9 , 7, 9 , 3 , 2 , 3 , 8 , 4 , 6 , 2 , 6 , 4 , 3 , 3 , 8 , 3 , 2 , 7, 9

The length of a digital subsequence is the number of integers it contains, not
the number of digits; the preceding example has length 12. As usual, a digital
subsequence is increasing if each number is larger than its predecessor.

Describe and analyze an efficient algorithm to compute the longest
increasing digital subsequence of D. [Hint: Be careful about your com¬
putational assumptions. How long does it take to compare two k-digit
numbers?]

For full credit, your algorithm should run in 0(n 4 ) time; faster algorithms
are worth extra credit. The fastest algorithm I know for this problem runs in
0 (n 3 ^ 2 log n) time; achieving this bound requires several tricks, both in the
design of the algorithm and in its analysis, but nothing outside the scope of
this class . 23

* 33 . Consider the following variant of the classical Tower of Hanoi problem. As
usual, there are n disks with distinct sizes, placed on three pegs numbered
0, 1, and 2. Initially, all n disks are on peg 0, sorted by size from smallest
on top to largest on bottom. Our goal is to move all the disks to peg 2. In
a single step, we can move the highest disk on any peg to a different peg,

23 With more advanced techniques, I believe the running time can be reduced to
0(n 3/2 loglog n), but I haven’t worked through the details.

138

Exercises

provided we satisfy two constraints. First, we must never place a smaller
disk on top of a larger disk. Second—and this is the non-standard part— we
must never move a disk directly from peg 0 to peg 2.

Describe and analyze an algorithm to compute the exact number of
moves required to move all n disks from peg 0 to peg 2 , subject to the
stated restrictions. For full credit, your algorithm should use only O(logn)
arithmetic operations in the worst case. For the sake of analysis, assume
that adding or multiplying two k-digit numbers requires 0(/c) time. [Hint:
Matrices!]

Splitting Sequences/Arrays

34 . A basic arithmetic expression is composed of characters from the set
{1, +, x} and parentheses. Almost every integer can be represented by more
than one basic arithmetic expression. For example, all of the following basic
arithmetic expression represent the integer 14 :

l+l+l+l+l+l+l+l+l+l+l+l+l+l+l+l
((1 + 1) x (1 + 1 + 1 + 1 + 1)) + ((1 + 1) x (1 + 1))

(l + l)x(l + l + l + l + l + l + l)

(l + l)x(((l + l + l)x(l + l)) + l)

Describe and analyze an algorithm to compute, given an integer n as input,
the minimum number of Is in a basic arithmetic expression whose value is
equal to n. The number of parentheses doesn’t matter, just the number of
Is. For example, when n = 14, your algorithm should return 8 , for the final
expression above. The running time of your algorithm should be bounded
by a small polynomial function of n.

35 . Suppose you are given a sequence of integers separated by + and — signs;
for example:

1 + 3 — 2 — 5 + 1 — 6 + 7

You can change the value of this expression by adding parentheses in
different places. For example:

1 + 3 — 2 — 5 + 1 — 6 + 7 = —1
(1 + 3 — (2 — 5)) + (1 — 6) + 7 = 9
(1+ (3-2))-(5 + 1)-(6 + 7) = -17

Describe and analyze an algorithm to compute, given a list of integers
separated by + and — signs, the maximum possible value the expression

139

3. Dynamic Programming

can take by adding parentheses. Parentheses must be used only to group
additions and subtractions; in particular, do not use them to create implicit
multiplication as in 1 + 3(—2)(—5) + 1 — 6 + 7 = 33.

36 . Suppose you are given a sequence of integers separated by + and x signs;
for example:

1 + 3x2x 0+1x 6 + 7

You can change the value of this expression by adding parentheses in
different places. For example:

(1 + (3 x 2)) x 0 + (1 x 6) + 7 = 13
((1 + (3 x 2 x 0) + 1) x 6 ) + 7 = 19
(1 + 3) x 2 x (0 + 1) x (6 + 7) = 208

(a) Describe and analyze an algorithm to compute the maximum possible
value the given expression can take by adding parentheses, assuming all
integers in the input are positive. [Hint: This is easy.]

(b) Describe and analyze an algorithm to compute the maximum possible
value the given expression can take by adding parentheses, assuming all
integers in the input are non-negative.

(c) Describe and analyze an algorithm to compute the maximum possible
value the given expression can take by adding parentheses, with no
restrictions on the input numbers.

Assume any arithmetic operation takes 0(1) time.

37 . After graduating from Shampoo-Banana University, you decide to interview
for a position at the Wall Street bank Long Live Boole. The managing
director of the bank, Eloob Egroeg, poses a ’solve-or-die’ problems to each
new employee, which they must solve within 24 hours. Those who fail to
solve the problem are fired immediately!

Entering the bank for the first time, you notice that the employee offices
are organized in a straight row, with a large T or F printed on the door of
each office. Furthermore, between each adjacent pair of offices, there is a
board marked by one of the symbols A, V, or ®. When you ask about these
arcane symbols, Eloob confirms that T and F represent the boolean values
True and False, and the symbols on the boards represent the standard
boolean operators And, Or, and Xor. He also explains that these letters
and symbols describe whether certain combinations of employees can work
together successfully. At the start of any new project, Eloob hierarchically
clusters his employees by adding parentheses to the sequence of symbols, to

140

Exercises

obtain an unambiguous boolean expression. The project is successful if this
parenthesized boolean expression evaluates to T.

For example, if the bank has three employees, and the sequence of
symbols on and between their doors is T A F ffi T, there is exactly one
successful parenthesization scheme: (T A (F © T)). However, if the list of
door symbols is F A T ® F, there is no way to add parentheses to make the
project successful.

Eloob finally poses your solve-or-die interview question: Describe an
algorithm to decide whether a given sequence of symbols can be parenthe¬
sized so that the resulting boolean expression evaluates to T. Your input is
an array S[0.. 2n], where S[i] € {T, F} when i is even, and S[i] € {V, A, ®}
when i is odd.

38 . Every year, as part of its annual meeting, the Antarctican Snail Lovers of
Upper Glacierville hold a Round Table Mating Race. Several high-quality
breeding snails are placed at the edge of a round table. The snails are
numbered in order around the table from 1 to n. During the race, each snail
wanders around the table, leaving a trail of slime behind it. The snails have
been specially trained never to fall off the edge of the table or to cross a
slime trail, even their own. If two snails meet, they are declared a breeding
pair, removed from the table, and whisked away to a romantic hole in the
ground to make little baby snails. Note that some snails may never find a
mate, even if the race goes on forever.

1

Figure 3.6. The end of a typical Antarctican SLUG race. Snails 6 and 8 never find mates. The organizers
must pay M[3,4] + M[2,5] + M[l, 7].

For every pair of snails, the Antarctican SLUG race organizers have
posted a monetary reward, to be paid to the owners if that pair of snails
meets during the Mating Race. Specifically, there is a two-dimensional

141

3. Dynamic Programming

array M[1.. n, 1.. n] posted on the wall behind the Round Table, where
M[i,j ] = M[j,i] is the reward to be paid if snails t and j meet.

Describe and analyze an algorithm to compute the maximum total
reward that the organizers could be forced to pay, given the array M as
input.

39 . You have mined a large slab of marble from a quarry. For simplicity, suppose
the marble slab is a rectangle measuring n inches in height and m inches
in width. You want to cut the slab into smaller rectangles of various sizes—
some for kitchen counter tops, some for large sculpture projects, others for
memorial headstones. You have a marble saw that can make either horizontal
or vertical cuts across any rectangular slab. At any time, you can query the
spot price P[x,y ] of an x-inch by y-inch marble rectangle, for any positive
integers x and y. These prices depend on customer demand, and people
who buy marble counter tops are weird, so don’t make any assumptions
about them; in particular, larger rectangles may have significantly smaller
spot prices. Given the array of spot prices and the integers m and n as input,
describe an algorithm to compute how to subdivide annxm marble slab to
maximize your profit.

40 . This problem asks you to design efficient algorithms to construct optimal
binary search trees that satisfy additional balance constraints. Your input
consists of a sorted array A[1.. n] of search keys and an array / [I.. n] of
frequency counts, where f[i] is the number of searches for A[i], This is
exactly the same cost function as described in Section 3 . 9 . But now your
task is to compute an optimal tree that satisfies some additional constraints.

(a) AVL trees were the earliest self-balancing balanced binary search trees,
first described in 1962 by Georgy Adelson-Velsky and Evgenii Landis. An
AVL tree is a binary search tree where for every node v, the height of
the left subtree of v and the height of the right subtree of v differ by at
most one.

Describe and analyze an algorithm to construct an optimal AVL tree
for a given set of search keys and frequencies.

(b) Symmetric binary B-trees are another self-balancing binary trees, first
described by Rudolf Bayer; these are better known by the name red-
black trees, after a somewhat simpler reformulation by Leo Guibas and
Bob Sedgwick in 1978 . A red-black tree is a binary search tree with the
following additional constraints:

• Every node is either red or black.

• Every red node has a black parent.

• Every root-to-leaf path contains the same number of black nodes.

142

Exercises

Describe and analyze an algorithm to construct an optimal red-black
tree for a given set of search keys and frequencies.

(c) AA trees were proposed by proposed by Arne Andersson in 1993 and
slightly simplified (and named) by Mark Allen Weiss in 2000 . AA
trees are also known as left-leaning red-black trees, after a symmetric
reformulation (with different rebalancing algorithms) by Bob Sedgewick
in 2006 . An AA-tree is a red-black tree with one additional constraint:

• No left child is red . 24

Describe and analyze an algorithm to construct an optimal AA-tree for a
given set of search keys and frequencies.

41 . Suppose you are given an m x n bitmap as an array M[1.. m, 1.. n] of Os
and Is. A solid block in M is a subarray of the form M[i .. i',j .. i 7 ] in which
all bits are equal. Suppose you want to decompose M into as few disjoint
blocks as possible.

One natural recursive partitioning strategy is called a guillotine sub¬
division. If the entire bitmap M is a solid block, there is nothing to do.
Otherwise, we cut M into two smaller bitmaps along a horizontal or vertical
line, and then recursively decompose the two smaller bitmaps into solid
blocks.

Any guillotine subdivision can be represented as a binary tree, where
each internal node stores the position and orientation of a cut, and each
leaf stores a single but 0 or 1 indicting the contents of the corresponding
block. The size of a guillotine subdivision is the number of leaves in the
corresponding binary tree (that is, the final number of solid blocks), and the
depth of a guillotine subdivision is the depth of the corresponding binary
tree.

(a) Describe and analyze an algorithm to compute a guillotine subdivision
of M of minimum possible size.

(b) Show that a guillotine subdivision does not always yield a partition into
the smallest number of solid blocks.

(c) Describe and analyze an algorithm to compute a guillotine subdivision
for M with the smallest possible depth.

(d) Describe and analyze an algorithm to determine M[i,j], given the tree
representing a guillotine decomposition for M and two indices i and j.

24 Sedgwick’s reformulation requires that no right child is red. Whatever. Andersson and
Sedgewick are all strangely silent about whether to measure angles clockwise or counterclockwise,
whether Pluto is a planet, whether “lower rank” means “better” or “worse”, or whether to fight a
hundred duck-sized horses or a single horse-sized duck.

M3

3. Dynamic Programming

Figure 3.7. A guillotine subdivision with size 8 and depth 5.

(e) Define the depth of a pixel M[i,j ] in a guillotine subdivision to be the
depth of the leaf that contains that pixel. Describe and analyze an
algorithm to compute a guillotine subdivision for M such that the sum
of the depths of the pixels as small as possible.

Cf) Describe and analyze an algorithm to compute a guillotine subdivision
for M such that the sum of the depths of the black pixels as small as
possible.

* 42 . Congratulations! You’ve been hired by the giant online bookstore DeNile
(“Not just a river in Egypt!”) to optimize their warehouse robots. Each book
that DeNile sells has a unique ISBN (International Standard Book Number),
which is just a numerical value. Each of DeNile’s warehouses contains a
long row of bins, each containing multiple copies of a single book. These
bins are arranged in sorted order by ISBN; each bin’s ISBN is printed on the
front of the bin in machine-readable form. Books are retrieved from these
bins by robots, which run along rails parallel to the row of bins.

DeNile does not maintain a list of which bins contain which ISBN
numbers; that would be too simple! Instead, to retrieve a desired book,
the robot must first find that book’s bin using a binary search. Because the
search requires physical motion by the robot, we can no longer assume that
each step of the binary search requires 0(1) time. Specifically:

• The robot always starts at the “Oth bin” (where the books are loaded
into boxes to ship to customers).

144

Exercises

• Moving the robot from the ith bin to the jth bin requires a\i—j\ seconds
for some constant a.

• The robot must be directly in front of a bin in order to read that bin’s
ISBN. Reading an ISBN requires /3 seconds, for some constant /3.

• Reversing the robot’s direction of motion (from increasing to decreasing
or vice versa) requires y additional seconds, for some constant y.

• When the robot finds the target bin, it extracts one book from that bin
and returns to “the Oth bin”.

Design and analyze an algorithm to compute a binary search tree over the
bins that minimizes the total time the robot spends searching for books.
Your input is an array/[I.. n] of integers, where / [t] is the number of times
that the robot will be asked to retrieve a book from the ith bin, along with
the time parameters a, 15, and y.

* 43 . A standard method to improve the cache performance of search trees is to
pack more search keys and subtrees into each node. A B-tree is a rooted
tree in which each internal node stores up to B keys and pointers to up to
B + 1 children, each the root of a smaller B-tree. Specifically, each node v
stores three fields:

• a positive integer v.d <B,

• a sorted array v.key[ 1 .. v.d], and

• an array v.child[ 0 .. v.d] of child pointers.

In particular, the number of child pointers is always exactly one more than
the number of keys . 25

Each pointer v.child[i] is either Null or a pointer to the root of a B-
tree whose keys are all larger than v.key[i] and smaller than v.key[i + 1 ].
In particular, all keys in the leftmost subtree v.child[0] are smaller than
v.key[ 1 ], and all keys in the rightmost subtree v.child[v.d ] are larger than
v.key[v.d ].

Intuitively, you should have the following picture in mind:

[ ? < key[ 1 ] < t < fcey[2] < • ■ • • • < key[d ] < • ]

¥ ¥ ¥ ¥ ¥

To T l T 2 ■■■ r d _! T d

25 Normally, B-trees are required to satisfy two additional constraints, which guarantee a
worst-case search cost of 0(log B n): Every leaf must have exactly the same depth, and every node
except possibly the root must contain at least B/2 keys. However, in this problem, we are not
interested in optimizing the worst-case search cost, but rather the total cost of a sequence of
searches, so we will not impose these additional constraints.

145

3. Dynamic Programming

Here Tj is the subtree pointed to by child[i].

The cost of searching for a key x in a B-tree is the number of nodes in
the path from the root to the node containing x as one of its keys. A 1-tree
is just a standard binary search tree.

Fix an arbitrary positive integer B > 0. (I suggest B = 8 .) Suppose
your are given a sorted array A[l, ..., n] of search keys and a corresponding
array F[l,..., n] of frequency counts, where F[i] is the number of times
that we will search for A[i], Your task is to describe and analyze an efficient
algorithm to find a B-tree that minimizes the total cost of searching for the
given keys with the given frequencies.

(a) Describe a polynomial-time algorithm for the special case B = 2.

(b) Describe an algorithm for arbitrary B that runs in 0(n l:+c ) time for some
fixed integer c.

*(c) Describe an algorithm for arbitrary B that runs in 0(n c ) time for some
fixed integer c that does not depend on B.

44 . A string w of parentheses ( and ) and brackets [ and ] is balanced if it
satisfies one of the following conditions:

• w is the empty string.

• w = (x) for some balanced string x

• w = [x] for some balanced string x

• w = xy for some balanced strings x and y
For example, the string

w= ([()][]()) COOK)

is balanced, because w = xy, where

x = ([()][]()) and y = [ ()()]()•

(a) Describe and analyze an algorithm to determine whether a given string
of parentheses and brackets is balanced.

(b) Describe and analyze an algorithm to compute the length of a longest
balanced subsequence of a given string of parentheses and brackets.

(c) Describe and analyze an algorithm to compute the length of a shortest
balanced supersequence of a given string of parentheses and brackets.

(d) Describe and analyze an algorithm to compute the minimum edit distance
from a given string of parentheses and brackets to a balanced string of
parentheses and brackets.

146

Exercises

v (e) Describe and analyze an algorithm to compute the longest common
balanced subsequence of two given strings of parentheses and brackets.

v (f) Describe and analyze an algorithm to compute the longest palindromic
balanced subsequence of a given string of parentheses and brackets.

r (g) Describe and analyze an algorithm to compute the longest common
palindromic balanced subsequence (whew!) of two given strings of
parentheses and brackets.

For each problem, your input is an array w[l.. n], where w[i] €{(,),[,]}
for every index i. (You may prefer to use different symbols instead of
parentheses and brackets—for example, L, R, l, r or <, >, ◄, ►—but please
tell your grader what symbols you’re using!)

v 45 . Congratulations! Your research team has just been awarded a $soM multi¬
year project, jointly funded by DARPA, Google, and McDonald’s, to produce
DWIM: The first compiler to read programmers’ minds! Your proposal and
your numerous press releases all promise that DWIM will automatically
correct errors in any given piece of code, while modifying that code as little
as possible. Unfortunately, now it’s time to start actually making the damn
thing work.

As a warmup exercise, you decide to tackle the following necessary
subproblem. Recall that the edit distance between two strings is the minimum
number of single-character insertions, deletions, and replacements required
to transform one string into the other. An arithmetic expression is a string w
such that

• w is a string of one or more decimal digits,

• w = (x) for some arithmetic expression x, or

• w = x o y for some arithmetic expressions x and y and some binary
operator o.

Suppose you are given a string of tokens from the alphabet {#, o, (,)},
where # represents a decimal digit and o represents a binary operator.
Describe and analyze an algorithm to compute the minimum edit distance
from the given string to an arithmetic expression.

46 . Ribonucleic acid (RNA) molecules are long chains of millions of nucleotides
or bases of four different types: adenine (A), cytosine (C), guanine (G), and
uracil (U). The sequence of an RNA molecule is a string b[l..n], where
each character b[i] e {A, C, G, U} corresponds to a base. In addition to the
chemical bonds between adjacent bases in the sequence, hydrogen bonds
can form between certain pairs of bases. The set of bonded base pairs is
called the secondary structure of the RNA molecule.

147

3. Dynamic Programming

We say that two base pairs (i, j) and (i',j') with i < j and i' < j'
overlap if i < i' < j < / or i' < i < j' < j. In practice, most base pairs are
non-overlapping. Overlapping base pairs create so-called pseudoknots in the
secondary structure, which are essential for some RNA functions, but are
more difficult to predict.

Suppose we want to predict the best possible secondary structure for
a given RNA sequence. We will adopt a drastically simplified model of
secondary structure:

• Each base can bond with at most one other base.

• Only A-U pairs and C-G pairs can bond.

• Pairs of the form (i, i + 1) and (i, i + 2) cannot bond.

• Bonded base pairs cannot overlap.

The last (and least realistic) restriction allows us to visualize RNA secondary
structure as a sort of fat tree, as shown below.

/G' C 'u'.,

A V:
\>/c' G 'a.

c

/

U-A-C-U-C-A-U

I I I I I I I

..-"r>A-U-G-A-G-U-A

/G A:--

v x

C \.^ U ' r A u —A'

^C\

'A' S'

A-U
I I

G-C

U-A

I I

U-A

\ ••

\ c;

r

U-U-A-C-

I I I I
A-A-U-G-

>A- :

' A; c\

'c'i

/ :

-u 4 -

; /

■A

V K g _ g -

Figure 3.8. Example RNA secondary structure with 21 bonded base pairs, indicated by heavy red lines.
Gaps are indicated by dotted curves. This structure has score 2 2 + 2 2 + 8 2 +1 2 + 7 2 + 4 2 + 7 2 = 187.

(a) Describe and analyze an algorithm that computes the maximum possible
number of bonded base pairs in a secondary structure for a given RNA
sequence.

(b) A gap in a secondary structure is a maximal substring of unpaired bases.
Large gaps lead to chemical instabilities, so secondary structures with
smaller gaps are more likely. To account for this preference, let’s define
the score of a secondary structure to be the sum of the squares of the gap
lengths; see Figure 3 . 8 . (This score function is utterly fictional; real RNA
structure prediction requires much more complicated scoring functions.)

Describe and analyze an algorithm that computes the minimum
possible score of a secondary structure for a given RNA sequence.

148

Exercises

* 47 . (a) Describe and analyze an efficient algorithm to determine, given a string
w and a regular expression R, whether w e L(R).

(b) Generalized regular expressions allow the binary operator n (intersection)
and the unary operator -> (complement), in addition to the usual
• (concatenation), + (or), and * (Kleene closure) operators. NFA
constructions and Kleene’s theorem imply that any generalized regular
expression E represents a regular language L(E).

Describe and analyze an efficient algorithm to determine, given a
string w and a generalized regular expression E, whether w € L(E).

In both problems, assume that you are actually given a parse tree for the
(generalized) regular expression, not just a string.

Trees and Subtrees

48 . You’ve just been appointed as the new organizer of Giggle, Inc.’s annual
mandatory holiday party. The employees at Giggle are organized into a
strict hierarchy, that is, a tree with the company president at the root. The
all-knowing oracles in Human Resources have assigned a real number to
each employee measuring how “fun” the employee is. In order to keep
things social, there is one restriction on the guest list: an employee cannot
attend the party if their immediate supervisor is also present. On the other
hand, the president of the company must attend the party, even though she
has a negative fun rating; it’s her company, after all. Give an algorithm that
makes a guest list for the party that maximizes the sum of the “fun” ratings
of the guests.

49 . Since so few people came to last year’s holiday party, the president of Giggle,
Inc. decides to give each employee a present instead this year. Specifically,
each employee must receive on the three gifts: ( 1 ) an all-expenses-paid six-
week vacation anywhere in the world, ( 2 ) an all-the-pancakes-you-can-eat
breakfast for two at Jumping Jack Flash’s Flapjack Stack Shack, or ( 3 ) a
burning paper bag full of dog poop. Corporate regulations prohibit any
employee from receiving exactly the same gift as his/her direct supervisor.
Any employee who receives a better gift than his/her direct supervisor will
almost certainly be fired in a fit of jealousy.

As Giggle, Inc.’s official party czar, it’s your job to decide which gift each
employee receives. Describe an algorithm to distribute gifts so that the
minimum number of people are fired. Yes, you may send the president a
flaming bag of dog poop.

More formally, you are given a rooted tree T, representing the company
hierarchy, and you want to label the nodes of T with integers 1, 2, or 3, so

149

3. Dynamic Programming

that every node has a different label from its parent. The cost of an labeling
is the number of nodes with smaller labels than their parents. See Figure
3.9 for an example. Describe and analyze an algorithm to compute the
minimum-cost labeling of T.

Figure 3.9. A tree labeling with cost 9 . The nine bold nodes have smaller labels than their parents This
is not the optimal labeling for this tree.

50 . After the Flaming Dog Poop Holiday Debacle, you were strongly encouraged
to seek other employment, and so you left Giggle for rival company Twitbook.
Unfortunately, the new president of Twitbook just decided to imitate Giggle
by throwing her own holiday party, and in light of your past experience,
appointed you as the official party organizer. The president demands that
you invite exactly k employees, including the president herself, and everyone
who is invited is required to attend. Yeah, that’ll be fun.

Just like at Giggle, employees at Twitbook are organized into a strict
hierarchy: a tree with the company president at the root. The all-knowing
oracles in Human Resources have assigned a real number to each employee
indicating the awkwardness of inviting both that employee and their imme¬
diate supervisor; a negative value indicates that the employee and their
supervisor actually like each other. Your goal is to choose a subset of
exactly k employees to invite, so that the total awkwardness of the resulting
party is as small as possible. For example, if the guest list does not include
both an employee and their immediate supervisor, the total awkwardness
is zero. The input to your algorithm is the tree T, the integer k, and the
awkwardness of each node in T.

(a) Describe an algorithm that computes the total awkwardness of the least
awkward subset of k employees, assuming the company hierarchy is
described by a binary tree. That is, assume that each employee directly
supervises at most two others.

r (b) Describe an algorithm that computes the total awkwardness of the least
awkward subset of k employees, with no restrictions on the company
hierarchy.

150

Exercises

51 . Suppose we need to broadcast a message to all the nodes in a rooted tree.
Initially, only the root node knows the message. In a single round, any node
that knows the message can forward it to at most one of its children. See
Figure 3.10 for an example.

(a) Design an algorithm to compute the minimum number of rounds required
to broadcast the message to all nodes in a binary tree.

4 (b) Design an algorithm to compute the minimum number of rounds required
to broadcast the message to all nodes in an arbitrary rooted tree. [Hint:
You may find techniques in the next chapter useful to prove your
algorithm is correct, even though it’s not a greedy algorithm.]

Figure 3.10. A message being distributed through a tree in five rounds.

52 . One day, Alex got tired of climbing in a gym and decided to take a very large
group of climber friends outside to climb. The climbing area where they
went, had a huge wide boulder, not very tall, with various marked hand
and foot holds. Alex quickly determined an “allowed” set of moves that her
group of friends can perform to get from one hold to another.

The overall system of holds can be described by a rooted tree T with n
vertices, where each vertex corresponds to a hold and each edge corresponds
to an allowed move between holds. The climbing paths converge as they go
up the boulder, leading to a unique hold at the summit, represented by the
root of T . 26

Alex and her friends (who are all excellent climbers) decided to play a
game, where as many climbers as possible are simultaneously on the boulder
and each climber needs to perform a sequence of exactly k moves. Each
climber can choose an arbitrary hold to start from, and all moves must move
away from the ground. Thus, each climber traces out a path of k edges
in the tree T, all directed toward the root. However, no two climbers are
allowed to touch the same hold; the paths followed by different climbers
cannot intersect at all.

Describe and analyze an efficient algorithm to compute the maximum
number of climbers that can play this game. More formally, you are given a
rooted tree T and an integer k, and you want to find the largest possible

26 Q: Why do computer science professors think trees have their roots at the top?

A: Because they’ve never been outside!

151

3. Dynamic Programming

number of disjoint paths in T, where each path has length k. Do not assume
that T is a binary tree. For example, given the tree T below and k = 3 as
input, your algorithm should return the integer 8 .

Figure 3.11. Seven disjoint paths of length k = 3. This is not the largest such set of paths in this tree.

53 . Let T be a rooted binary tree with n vertices, and let k < n be a positive
integer. We would like to mark k vertices in T so that every vertex has a
nearby marked ancestor. More formally, we define the clustering cost of any
subset K of vertices as

cost(fC) = max cost(v,K),

V

where the maximum is taken over all vertices v in the tree, and cost(v,K ) is
the distance from v to its nearest ancestor in K:

{ 0 iiveK

00 if v is the root of T and v ^ K

1 + cost(parent(v )) otherwise

In particular, cost(K ) = 00 if K excludes the root of T.

Figure 3.12. A subset of five vertices in a binary tree, with clustering cost 3 .

152

Exercises

*(a) Describe a dynamic programming algorithm to compute, given the tree
T and an integer k, the minimum clustering cost of any subset of k
vertices in T. For full credit, your algorithm should run in 0(n 2 k 2 ) time.

(b) Describe a dynamic programming algorithm to compute, given the tree
T and an integer r, the size of the smallest subset of vertices whose
clustering cost is at most r. For full credit, your algorithm should run in
O(nr) time.

(c) Show that your solution for part (b) implies an algorithm for part (a)
that runs in 0 (n 2 logn) time.

54 . This question asks you to find efficient algorithms to compute the largest
common rooted subtree of two given rooted trees. Recall that a rooted
tree is a connected acyclic graph with a designated node called the root.
A rooted subtree of a rooted tree consists of an arbitrary node and all its
descendants. The precise definition of “common” depends on which pairs of
rooted trees we consider isomorphic.

(a) Recall that a binary tree is a rooted tree in which every node has a
(possibly empty) left subtree and a (possibly empty) right subtree. Two
binary trees are isomorphic if and only if they are both empty, or their
left subtrees are isomorphic and their right subtrees are isomorphic.
Describe an algorithm to find the largest common binary subtree of two
given binary trees.

Figure 3.13. Two binary trees, with their largest common (rooted) subtree emphasized.

(b) In an ordered rooted tree, each node has a sequence of children, which
are the roots of ordered rooted subtrees. Two ordered rooted trees are
isomorphic if they are both empty, or if their ith subtrees are isomorphic
for every index i. Describe an algorithm to find the largest common
ordered subtree of two ordered trees T 1 and T 2 .

♦ v (c) In an unordered rooted tree, each node has an unordered set of children,
which are the roots of unordered rooted subtrees. Two unordered rooted
trees are isomorphic if they are both empty, or the subtrees of each root
can be ordered so that their ith subtrees are isomorphic for every index i.
Describe an algorithm to find the largest common unordered subtree of
two unordered trees T 1 and T 2 .

153

3. Dynamic Programming

55 . This question asks you to find efficient algorithms to compute optimal
subtrees in unrooted trees—connected acyclic undirected graphs. A subtree
of an unrooted tree is any connected subgraph.

(a) Suppose you are given an unrooted tree T with weights on its edges,
which may be positive, negative, or zero. Describe an algorithm to find
a path in T with maximum total weight.

(b) Suppose you are given an unrooted tree T with weights on its vertices,
which may be positive, negative, or zero. Describe an algorithm to find
a subtree of T with maximum total weight. [This was a 2016 Google
interview question.]

(c) Let T 1 and T 2 be arbitrary ordered unrooted trees, meaning that the
neighbors of every node have a well-defined cyclic order. Describe an
algorithm to find the largest common ordered subtree of and T 2 .

* r (d) Let T 1 and T 2 be arbitrary unordered unrooted trees. Describe an
algorithm to find the largest common unordered subtree of Tj and T 2 .

56 . Rooted minors of rooted trees are a natural generalization of subsequences.
A rooted minor of a rooted tree T is any tree obtained by contracting one or
more edges. When we contract an edge u-»v, where u is the parent of v, the
children of v become new children of u and then v is deleted. In particular,
the root of T is also the root of every rooted minor of T.

Figure 3.14. A rooted tree and one of its rooted minors.

(a) Let T be a rooted tree with labeled nodes. We say that T is boring
if, for each node x, all children of x have the same label; children of
different nodes may have different labels. Describe an algorithm to find
the largest boring rooted minor of a given labeled rooted tree.

(b) Suppose we are given a rooted tree T whose nodes are labeled with
numbers. Describe an algorithm to find the largest heap-ordered rooted
minor of T. That is, your algorithm should return the largest rooted
minor M such that every node in M has a smaller label than its children
in M.

154

Exercises

(c) Suppose we are given a binary tree T whose nodes are labeled with
numbers. Describe an algorithm to find the largest binary-search-ordered
rooted minor of T. That is, your algorithm should return a rooted minor
M such that every node in M has at most two children, and an inorder
traversal of M is an increasing subsequence of an inorder traversal of T.

(d) Recall that a rooted tree is ordered if the children of each node have a
well-defined left-to-right order. Describe an algorithm to find the largest
binary-search-ordered minor of an arbitrary ordered tree T whose nodes
are labeled with numbers. Again, the left-to-right order of nodes in M
should be consistent with their order in T.

r (e) Describe an algorithm to find the largest common ordered rooted minor
of two ordered labeled rooted trees.

4 *(f) Describe an algorithm to find the largest common unordered rooted
minor of two unordered labeled rooted trees. [Hint: Combine dynamic
programming with maximum flows.]

155

The point is, ladies and gentleman, greed is good. Greed works, greed is right.

Greed clarifies, cuts through, and captures the essence of the evolutionary spirit.
Greed in all its forms, greed for life, money, love, knowledge
has marked the upward surge in mankind.

And greed-mark my words—will save not only Teldar Paper
but the other malfunctioning corporation called the USA.

- Gordon Gekko [Michael Douglas], Wall Street (1987)

There is always an easy solution to every human problem-
neat, plausible, and wrong.

— H. L. Mencken, "The Divine Afflatus",
New York Evening Mail (November 16,1917)

Greedy Algorithms

4.1 Storing Files on Tape

Suppose we have a set of rt files that we want to store on magnetic tape. 1 In the
future, users will want to read those files from the tape. Reading a file from
tape isn’t like reading a file from disk; first we have to fast-forward past all the
other files, and that takes a significant amount of time. Let L[1.. n] be an array
listing the lengths of each file; specifically, file i has length L[i], If the files are
stored in order from 1 to n, then the cost of accessing the kth file is

k

cost(k) = ^L[i],

i=1

’Readers who are tempted to object that magnetic tape has been obsolete for decades
are cordially invited to tour your nearest supercomputer facility; ask to see the tape robots.
Alternatively, consider filing a sequence of books on a library bookshelf. You know, those strange
brick-like objects made of dead trees and ink?

157

4. Greedy Algorithms

The cost reflects the fact that before we read file k we must first scan past all the
earlier files on the tape. If we assume for the moment that each file is equally
likely to be accessed, then the expected cost of searching for a random file is

If we change the order of the files on the tape, we change the cost of accessing
the files; some files become more expensive to read, but others become cheaper.
Different file orders are likely to result in different expected costs. Specifically,
let Tt(i) denote the index of the file stored at position i on the tape. Then the
expected cost of the permutation n is

fc=1 ;=1

Which order should we use if we want the expected cost to be as small as
possible? The answer is intuitively clear; we should store the files in order from
shortest to longest. So let’s prove this.

Lemma 4 . 1 . E[cost(7i)] is minimized when L[tt(i)] < L[n(i + 1)] for all i.

Proof: Suppose L[7i(i)] > L[tr(i + 1)] for some i. To simplify notation, let
a = tr(i) and b = n{i + 1). If we swap files a and b, then the cost of accessing
a increases by L[b], and the cost of accessing b decreases by L[a]. Overall,
the swap changes the expected cost by (L[b] — L[a])/n. But this change is an
improvement, because L[b ] < L[a]. Thus, if the files are out of order, we can
improve the expected cost by swapping some mis-ordered adjacent pair. □

This example gives us our first greedy algorithm. To minimize the total
expected cost of accessing the files, we put the file that is cheapest to access
first, and then recursively write everything else; no backtracking, no dynamic
programming, just make the best local choice and blindly plow ahead. If we
use an efficient sorting algorithm, the running time is clearly 0(n log n), plus
the time required to actually write the files. To show that the greedy algorithm
is actually correct, we prove that the output of any other algorithm can be
improved by some sort of exchange

Let’s generalize this idea further. Suppose we are also given an array F[1.. n]
of access frequencies for each file; file i will be accessed exactly F[t] times over
the lifetime of the tape. Now the total cost of accessing all the files on the tape is

n f k \ n k

Ecost(n) = £ FMk)] ■ 2 LMi)] = XS (W fc )] ■ L MQ1).
k =1 V i=1 J k =1 ;=1

158

4.2. Scheduling Classes

Now what order should store the files if we want to minimize the total cost?
(This question is similar in spirit to the optimal binary search tree problem,
but the target data structure and the cost function are both different, so the
algorithm must be different, too.)

We’ve already proved that if all the frequencies are equal, then we should
sort the files by increasing size. If the frequencies are all different but the file
lengths L[i] are all equal, then intuitively, we should sort the files by decreasing
access frequency, with the most-accessed file first. In fact, this is not hard to
prove by modifying the proof of Lemma 4.1. But what if the sizes and the
frequencies both vary? In this case, we should sort the files by the ratio L/F.

Lemma 4 . 2 . T.cost{n) is minimized when

b[ 7 t(Q] < L[n(i + 1)]
f[7i(i)] - F[n(i + 1)]

for all i.

Proof: Suppose L[n(i)]/F[n{i)] > L[n(i + 1 )]/F\n(i + i)] for some i. To
simplify notation, let a = 7t(i) and b = n(i + 1 ). If we swap files a and b, then
the cost of accessing a increases by L[b], and the cost of accessing b decreases
by L[a], Overall, the swap changes the total cost by L[b]F[a] — L[a]F[b], But
this change is an improvement, because

b[q] L[b]
F[a] F[b ]

L[b]F[a]-L[a]F[b] < 0 .

Thus, if any two adjacent files are out of order, we can improve the total cost by
swapping them. □

4.2 Scheduling Classes

The next example is slightly more complex. Suppose you decide to drop out
of computer science and change your major to Applied Chaos. The Applied
Chaos department offers all of its classes on the same day every week, called
“Soberday” by the students (but interestingly, not by the faculty). Every class
has a different start time and a different ending time: AC 101 (“Toilet Paper
Landscape Architecture”) starts at 10:27pm and ends at 11:51pm; AC 666
(“Immanentizing the Eschaton”) starts at 4:18pm and ends at 4:22pm, and so on.
In the interest of graduating as quickly as possible, you want to register for as
many classes as possible. (Applied Chaos classes don’t require any actual work.)
The university’s registration computer won’t let you register for overlapping
classes, and no one in the department knows how to override this “feature”.
Which classes should you take?

More formally, suppose you are given two arrays S[ 1 .. n] and F [ 1 .. n] listing
the start and finish times of each class; to be concrete, we can assume that
0 <S[i]<F[i]<M for each i, for some value M (for example, the number

159

4. Greedy Algorithms

of picoseconds in Soberday). Your task is to choose the largest possible subset
X e { 1 , 2 ,.. n} so that for any pair i,j el, either S[i] > F[j] or S[j] > F[i].
We can illustrate the problem by drawing each class as a rectangle whose left
and right x-coordinates show the start and finish times. The goal is to find a
largest subset of rectangles that do not overlap vertically.

I 1 I I □

Figure 4.1. A maximum conflict-free schedule for a set of classes.

This problem has a fairly simple recursive solution, based on the observation
that either you take class 1 or you don’t. Let B denote the set of classes that end
before class 1 starts, and let A denote the set of classes that start after class 1
ends:

B := {i | 2 < i < n and F[i ] < S[l]}

A := {i | 2 < i < n and S[i] > F[l]}

If class 1 is in the optimal schedule, then so are the optimal schedules for B
and A, which we can find recursively. If not, we can find the optimal schedule
for { 2 , 3 ,..., n} recursively. So we should try both choices and take whichever
one gives the better schedule. Evaluating this recursive algorithm from the
bottom up gives us a dynamic programming algorithm that runs in 0(n 3 ) time.
I won’t bother to go through the details, because we can do better. 2

Intuitively, we’d like the first class to finish as early as possible, because that
leaves us with the largest number of remaining classes. This intuition suggests
the following simple greedy algorithm. Scan through the classes in order of
finish time; whenever you encounter a class that doesn’t conflict with your latest
class so far, take it! See Figure 4.2 for a visualization of the resulting greedy
schedule.

We can write the greedy algorithm somewhat more formally as follows.
(Hopefully the first line is understandable.) The algorithm clearly runs in
O(nlogn) time; the running time is dominated by the initial sort.

2 But you should still work out the details yourself. The dynamic programming algorithm

can be used to find the “best” schedule for several different definitions of “best”, but the greedy
algorithm I’m about to describe only works when “best” means “biggest”. Also, you can improve
the running time to 0(n 2 ) using a different recurrence.

160

4.2. Scheduling Classes

cd

1

□

c

1

1

1

1

1

Figure 4.2. The same classes sorted by finish times and the greedy schedule.

GreedyScheduleCST 1 ..nl,Fr 1 ..nl):

sort F and permute S to match
count <— 1
X[count ] <— 1
for i <— 2 to n

if S[i] > F[X[count]]
count <— count + 1
X[count ] <— i
return X[l ..count]

To prove that GreedySchedule actually computes the largest conflict-free
schedule, we use an exchange argument, similar to the one we used for tape
sorting. We are not claiming that the greedy schedule is the only maximal
schedule; there could be others. (Compare Figures 4.1 and 4.2!) All we can
claim is that at least one of the optimal schedules is the one produced by the
greedy algorithm.

Lemma 4 . 3 . At least one maximal conflict-free schedule includes the class that
finishes first.

Proof: Let / be the class that finishes first. Suppose we have a maximal
conflict-free schedule X that does not include /. Let g be the first class in X to
finish. Since / finishes before g does, / cannot conflict with any class in the set
S \ {g}. Thus, the schedule X' = X U {/} \ {g} is also conflict-free. Since X' has
the same size as X, it is also maximal. □

To finish the proof, we call on our old friend, induction.

4. Greedy Algorithms

Theorem 4 . 4 . The greedy schedule is an optimal schedule.

Proof: Let / be the class that finishes first, and let A be the subset of classes that
start after / finishes. The previous lemma implies that some optimal schedule
contains /, so the best schedule that contains / is an optimal schedule. The best
schedule that includes / must contain an optimal schedule for the classes that
do not conflict with /, that is, an optimal schedule for A. The greedy algorithm
chooses / and then, by the inductive hypothesis, computes an optimal schedule
of classes from A. □

The proof might be easier to understand if we unroll the induction slightly.

Proof: Let {g\, g 2 , ■ ■ ■, gk) be the sequence of classes chosen by the greedy
algorithm. Suppose we have a maximal conflict-free schedule

■5 = (gl, g2> ■ ■ ■ , gj- 1> c j, c j+1, • • • > c m)>

where class c } is different from the class g j chosen by the greedy algorithm. (We
could have j = 1, in which case this schedule starts with a non-greedy choice
(q.) By construction, the /th greedy choice g ; does not conflict with any earlier
class gi, g 2 ,..., gj-i, and because our schedule S is conflict-free, neither does c ; -.
Moreover, gj has the earliest finish time among all classes that don’t conflict
with the earlier classes; in particular, g, finishes before c } . It follows that g (
does not conflict with any of the later classes c J+1 ,..., c m . Thus, the schedule

S (gl; g2> • • • > gj— 1’ gj’ ^j+1’ • • •; ^m)’

is also conflict-free. (This argument is just a generalization of Lemma 4.3, which
considers the case j = 1.)

By induction, it now follows that there is an optimal schedule (g 1; g 2 ,..., gk,
c k+1 ,..., c m ) that includes every class chosen by the greedy algorithm. But this
is impossible unless k = m\ if some class c k+1 does not conflict with any of the
first k greedy classes, then the greedy algorithm would choose more than k
classes! □

4.3 General Structure

The basic structure of this correctness proof is exactly the same as for the
tape-sorting problem: an inductive exchange argument.

• Assume that there is an optimal solution that is different from the greedy
solution.

• Find the “first” difference between the two solutions.

162

44 - Huffman Codes

• Argue that we can exchange the optimal choice for the greedy choice without
making the solution worse (although the exchange might not make it better).

This argument implies by induction that some optimal solution contains the
entire greedy solution, and therefore equals the greedy solution. Sometimes, as
in the scheduling problem, an additional step is required to show no optimal
solution strictly improves the greedy solution.

4.4 Huffman Codes

A binary code assigns a string of os and is to each character in the alphabet. A
binary code is prefix-free if no code is a prefix of any other. (Confusingly, prefix-
free codes are also commonly called prefix codes.) 7-bit ASCII and Unicode’s
UTF-8 are both prefix-free binary codes. Morse code is a binary code with
symbols • and —, but it is not prefix-free, because the code for E (•) is a prefix
of the codes for I (• •), S (• • •), and H (• • • »). 3

Any prefix-free binary code can be visualized as a binary tree with the
encoded characters stored at the leaves. The code word for any symbol is given
by the path from the root to the corresponding leaf; 0 for left, 1 for right. Thus,
the length of any symbol’s codeword is the depth of the corresponding leaf in
the code tree. Although they are superficially similar, binary code trees are not
binary search trees; we don’t care at all about the order of symbols at the leaves.

Suppose we want to encode a message written in an n-character alphabet
so that the encoded message is as short as possible. Specifically, given an array
frequency counts / [I.. n], we want to compute a prefix-free binary code that
minimizes the total encoded length of the message:

n

X! fU]-depth(i ).

;=i

This is exactly the same cost function we considered for optimizing binary
search trees, but the optimization problem is different, because code trees are
not required to keep the keys in any particular order.

3 For this reason, Morse code is arguably better described as a prefix-free ternary code, with
three symbols: *, —, and pause. Alternatively, Morse code can be considered a prefix-free
binary code, with one beat of sound/light/current/high voltage/smoke/gas (■) and one beat
of silence/darkness/no current/low voltage/air/liquid (□) as the two symbols. Then each “dit”
is encoded as ■□, each “dah” as ■■■□, and each pause as □□. In standard Morse code, each
letter is followed by one pause, and each word is followed by two additional pauses; however,
□s at the end of the entire coded message are omitted. For example, the string “MORSE CODE” is
unambiguously encoded as the following bit string:

163

4. Greedy Algorithms

In 1951, as a PhD student at MIT, David Huffman developed the following
greedy algorithm to produce such an optimal code: 4

Huffman: Merge the two least frequent letters and recurse.

Huffman’s algorithm is best illustrated through an example. Suppose we want
to encode the following helpfully self-descriptive sentence, discovered by Lee
Sallows: 5

This sentence contains three a's, three c's, two d's, twenty-six e's, five fs, three
g's, eight h's, thirteen i's, two I's, sixteen n's, nine o's, six r's, twenty-seven s's,
twenty-two t's, two u's, five v's, eight w's, four x's, five y's, and only one z.

To keep things simple, let’s ignore the forty-four spaces, nineteen apostrophes,
nineteen commas, three hyphens, and only one period, and encode only the
letters:

THISSENTENCECONTAINSTHREEASTHREECSTWODSTWENTYSIXESFIVEFST

HREEGSEIGHTHSTHIRTEENISTWOLSSIXTEENNSNINEOSSIXRSTWENTYSEV

ENSSTWENTYTW0TSTW0USFIVEVSEIGHTWSF0URXSFIVEYSAND0NLY0NEZ 6

Here is the frequency table for Sallows’ sentence:

A

C

D

E

F

G

H

I

L

N

0

R

S

T

U

V

W

X

Y

Z

3

3

2

26

5

3

8

13

2

16

9

6

27

22

2

5

8

4

5

1

Huffman’s algorithm picks out the two least frequent letters, breaking ties
arbitrarily—in this case, say, Z and D —and merges them together into a single

4 Huffman was a student in an information theory class taught by Robert Fano, who was
a close colleague of Claude Shannon, the father of information theory. Fano and Shannon
had previously developed a different greedy algorithm for producing prefix codes—split the
frequency array into two subarrays as evenly as possible, and then recursively build a code for
each subarray—but these Fano-Shannon codes were known not to be optimal. Fano posed the
problem of finding an optimal prefix code to his class. Huffman decided to solve the problem as
a class project, instead of taking a final exam, not realizing that the problem was open, or that
Fano and Shannon had already tried and failed to solve it. After several months of fruitless effort,
Huffman eventually gave up and decided to take the final exam after all. As he was throwing his
notes in the trash, the solution dawned on him. Huffman would later describe the epiphany as
“the absolute lightning of sudden realization”.

5 This sentence was first reported by Alexander Dewdney in his October 1984 “Computer
Recreations” column in Scientific American. Sallows himself published the remarkable story of
its discovery in 1985, along with several other self-descriptive sentences; you can find Sallows’
paper on his web site. Frustrated with the slow progress of his code running on a VAX 11/780,
Sallows designed and built dedicated hardware to perform a brute-force search for self-descriptive
sentences with various flavor text (“This pangram has... ”, “This sentence contains exactly... ”,
and so on). Careful theoretical analysis limited the search space to just over six billion possibilities,
which his l-MHz Pangram Machine enumerated in just under two hours.

6 ... and he talked for forty-five minutes, and nobody understood a word that he said, but we
had fun fillin’ out the forms and playin’ with the pencils on the bench there.

164

44 - Huffman Codes

new character E with frequency 3 . This new character becomes an internal
node in the code tree we are constructing, with Z and D as its children; it
doesn’t matter which child is which. The algorithm then recursively constructs
a Huffman code for the new frequency table

A

C

E

F

G

H

I

L

N

0

R

S

T

U

V

W

X

Y

E

3

3

26

5

3

8

13

2

16

9

6

27

22

2

5

8

4

5

3

After 19 merges, all 20 letters have been merged together. The record of merges
gives us our code tree. The algorithm makes a number of arbitrary choices; as a
result, there are actually several different Huffman codes. One such Huffman
code is shown below; numbers in non-leaf nodes are frequencies for merged
characters. For example, the code for A is 110000 , and the code for S is 00 .

Encoding Sallows’ sentence with this particular Huffman code would yield a
bit string that starts like so:

1001 0100 1101 00 00 111 011 1001 111 011 110001 111 110001 10001 011 1001
THISSENTEN C E C ONT

Here is the list of costs for encoding each character in Sallows’ sentence, along
with that character’s contribution to the total length of the encoded sentence:

char

A

C

D

E

F

G

H

I

L

N

0

R

S

T

U

V

W

X

Y

Z

freq

3

3

2

26

5

3

8

13

2

16

9

6

27

22

2

5

8

4

5

1

depth

6

6

7

3

5

6

4

4

7

3

4

4

2

4

7

5

4

6

5

7

total

18

18

14

78

25

18

32

52

14

48

36

24

54

88

14

25

32

24

25

7

Altogether, the encoded message is 646 bits long. Different Huffman codes
encode the same characters differently, possibly with code words of different
length, but the overall length of the encoded message is the same for every
Huffman code: 646 bits.

165

4. Greedy Algorithms

Given the simple structure of Huffman’s algorithm, it’s rather surprising that
it produces an optimal prefix-free binary code. 7 Encoding Sallows’ sentence
using any prefix-free code requires at least 646 bits! Fortunately, the recursive
structure makes this claim easy to prove using an exchange argument, similar
to our earlier optimality proofs. We start by proving that the algorithm’s very
first choice is correct.

Lemma 4.5. Let x and y be the two least frequent characters (breaking ties
between equally frequent characters arbitrarily). There is an optimal code tree
in which x and y are siblings.

Proof: I’ll actually prove a stronger statement: There is an optimal code in
which x and y are siblings and have the largest depth of any leaf.

Let T be an optimal code tree, and suppose this tree has depth d. Because T
is a full binary tree, it has at least two leaves at depth d that are siblings. (Verify
this by induction!) Suppose those two leaves are not x and y, but some other
characters a and b.

Let T' be the code tree obtained by swapping x and a, and let A =
d—depth T (x). This swap increases the depth of x by A and decreases the depth
of a by A, so

cost(T ') = cost(T) + A • (/[x] —/[a]).

Our assumption that x is one of the two least frequent characters but a is not
implies /[x] < f[a], and our assumption that a has maximum depth implies
A > 0 . It follows that cost(T') < cost(T). On the other hand, T is an optimal
code tree, so we must also have cost(T') > cost(T). We conclude that T' is also
an optimal code tree.

Similarly, swapping y and b must give yet another optimal code tree. In this
final optimal code tree, x and y are maximum-depth siblings, as required. □

Now optimality is guaranteed by our dear friend the Recursion Fairy! Our
recursive argument relies on the following non-standard recursive definition:
A full binary tree is either a single node, or a full binary tree where some leaf has
been replaced by an internal node with two leaf children.

Theorem 4.6. Every Huffman code is an optimal prefix-free binary code.

Proof: If the message has only one or two distinct characters, the theorem is
trivial, so assume otherwise.

Let /[I.. n] be the original input frequencies, and assume without loss of
generality that /[l] and /[ 2 ] are the two smallest frequencies. To set up the

7 It was certainly surprising to both Huffman and Fano!

166

44 - Huffman Codes

recursive subproblem, define / [n + 1] = /[l]+/[2]. Our earlier exchange
argument implies that 1 and 2 are (deepest) siblings in some optimal code for
f[l..n].

Let T' be the Huffman tree for / [3 .. n + 1]; the inductive hypothesis implies
that T' is an optimal code tree for the smaller set of frequencies. To obtain
the final code tree T, we replace the leaf labeled n+1 with an internal node
with two children, labelled 1 and 2. I claim that T is optimal for the original
frequency array /[I.. n].

To prove this claim, we can express the cost of T in terms of the cost of T'
as follows. (In these equations, depth(i) denotes the depth of the leaf labelled i
in either T or T'; each leaf that appears in both T and T' has the same depth in
both trees.)

n

cost(T) = ^f[i] ■ depth(i)
i=1
n+1

= ^^f[i]-depth(i) + /[l] • depth(l) + /[2] • depth{ 2 )

;=3 - f[n + l]-depth(n + l)

= cost(T / ) + (/[l] + /[ 2 ])-depth(T) — f[n + 1 ] • (depth(T) — 1 )

= cost(T') + f[l] +/[2] + (/[l] +/[ 2 ] —f[n + 1]) • (depth(T) — 1)
= cost(T') + f[l]+f[ 2 ]

This equation implies that minimizing the cost of T is equivalent to minimizing
the cost of T'; in particular, attaching leaves labeled 1 and 2 to the leaf in T'
labeled n + 1 gives an optimal code tree for the original frequencies. □

To efficiently construct a Huffman code, we keep the characters in a priority
queue, using the character frequencies as priorities. We can represent the code
tree as three arrays of indices, listing the Left and flight children and the Parent
of each node. The leaves of the final code tree are nodes at indices 1 through
n, and the root is the node with index 2n — 1. Pseudocode for the algorithm is
shown in Figure 4 . 3 . BuildHuffman performs O(n) priority-queue operations:
exactly 2n — 1 Inserts and 2 n — 2 ExtractMins. If we implement the priority
queue as a standard binary heap, each of these operations requires O(logn)
time, and thus the entire algorithm runs in 0 (n logn) time.

Finally, simple algorithms to encode and decode messages using a fixed
Huffman code are shown in Figure 4 . 4 ; both algorithms run in O(m) time,
where m is the length of the encoded message.

167

4 . Greedy Algorithms

BuildHuffman( f |~1 nl):

for i <— 1 to n

L[i]<-0; R[i] <— 0
lNSERT(i,/[i])
for i <— n to 2n — 1

x <— ExtractMin( )
y <— ExtractMin( )

/[i]«-/M+/[y]

Insert(i,/[i])
L[i]*~x-, P[x] <— i

R[i]<-y ; P[y]«-i

P[2n — 1]«— 0

((find two rarest symbols))
((merge into a new symbol))
((update tree pointers))

Figure 4.3. Building a Huffman code.

HuffmanEncode(A[ 1.. k]):

HuffmanDecode(B[ 1.. ml):

m <— 1

k<— 1

for i <— 1 to k

v <— 2n — 1

HuffmanEncodeOne(A[i])

for i <— 1 to m

HuffmanEncodeOne(x):

if B[t] = 0
v <— L[v]

else

v <—R[v]

if L[v] = 0

if x < 2n — 1

HuffmanEncodeOne(P[x])

if x = L[P[x]]

B[m ] <— 0

else

A[k ]«- v

B[m ] <- 1

k^k + 1

m <— m + 1

v <— 2n — 1

Figure 4.4. Encoding and decoding algorithms for Huffman codes

4.5 Stable Matching

Every year, thousands of new doctors must obtain internships at hospitals around
the United States. During the first half of the 20th century, competition among
hospitals for the best doctors led to earlier and earlier offers of internships,
sometimes as early as the second year of medical school, along with tighter
deadlines for acceptance. In the 1940s, medical schools agreed not to release
information until a common date during their students’ fourth year. In response,
hospitals began demanding faster decisions. By 1950, hospitals would regularly
call doctors, offer them internships, and demand immediate responses. Interns
were forced to gamble if their third-choice hospital called first—accept and risk
losing a better opportunity later, or reject and risk having no position at all. 8

168

8 The American academic job market involves similar gambles, at least in computer science.
Some departments start making offers in February with two-week decision deadlines; other
departments don’t even start interviewing until March; MIT notoriously waits until May, when all

4 - 5 - Stable Matching

Finally, a central clearinghouse for internship assignments, now called the
National Resident Matching Program (NMRP), was established in the early 1950s.
Each year, doctors submit a ranked list of all hospitals where they would accept
an internship, and each hospital submits a ranked list of doctors they would
accept as interns. The NRMP then computes an matching between doctors
and hospitals that satisfies the following stability requirement. A matching is
unstable if there is a doctor a and hospital B that would be both happier with
each other than with their current match; that is,

• a is matched with some other hospital A, even though she prefers B.

• B is matched with some other doctor / 3 , even though they prefer a.

In this case, we call (a,B) an unstable pair for the matching. The goal of the
Resident Match is a stable matching, which is a matching with no unstable
pairs.

For simplicity, I’ll assume from now on that there are exactly the same
number of doctors and hospitals; each hospital offers exactly one internship;
each doctor ranks all hospitals and vice versa; and finally, there are no ties in
the doctors’ or hospitals’ rankings. * * * * * 9

Some Bad Ideas

At first glance, it is not clear that a stable matching always exists! Certainly
not every matching of doctors and hospitals is stable. Suppose there are three
doctors (Dr. Quincy, Dr. Rotwang, Dr. Shephard, represented by lower-case
letters) and three hospitals (Arkham Asylum, Bethlem Royal Hospital, and
County General Hospital, represented by upper-case letters), who rank each
other as follows:

q r s ABC

A C A r s q

CAB q q r

BBC s r s

The matching {Aq,Br,Cs} is unstable, because Arkham would rather hire
Dr. Rotwang than Dr. Quincy, and Dr. Rotwang would rather work at Arkham
than at Bedlam. (A, r) is an unstable pair for this matching.

its interviews are over, before making any faculty offers. Needless to say, the mishmash of offer

dates and decision deadlines causes tremendous stress, for candidates and departments alike.

For similar reasons, since 1965, most American universities have agreed to a common April 15

deadline for prospective graduate students to accept offers of financial support (and by extension,

offers of admission).

9 In reality, most hospitals offer multiple internships, each doctor ranks only a subset of the
hospitals and vice versa, and there are typically more internships than interested doctors. And
then it starts getting complicated.

169

4 . Greedy Algorithms

One might imagine using an incremental algorithm that starts with an
arbitrary matching, and then greedily performs exchanges to resolve instabilities.
Unfortunately, resolving one instability can new ones; in fact, this incremental
“improvement” can lead to an infinite loop. For example, if we start with our
earlier unstable matching {Aq,Br, Cs}, each of the following exchanges resolves
one unstable pair (indicated over the arrow), but the sequence of exchanges
leads back to the original matching: 10

{Aq,Br, Cs}

Alternatively, we might try the following multi-round greedy protocol. In
each round, every unmatched hospital makes an offer to their favorite unmatched
doctor, then every unmatched doctor with an offer accepts their favorite offer.
It’s not hard to prove that at least one new doctor-hospital pair is matched each
round, so the algorithm always ends with a matching. For the previous example
input, we already have a stable matching {Ar,Bs, Cq} at the end of the first
round! But consider the following input instead:

ABC
q q s

q r s

C A A
B C B
ABC

s r r
r s q

In the first round, Dr. Shephard accepts an offer from County, and Dr. Quincy
accepts an offer from Bedlam (rejecting Arkham’s offer), leaving only Dr.
Rotwang and Arkham unmatched. Thus, the protocol ends with the matching
{Ar,Bq, C.s} after two rounds. Unfortunately, this matching is unstable; Arkham
and Dr. Shephard prefer each other to their matches.

The Boston Pool and Gale-Shapley Algorithms

In 1952, the NRMP adopted the “Boston Pool” algorithm to assign interns, so
named because it had been previously used by a regional clearinghouse in the
Boston area. Ten years later, David Gale and Lloyd Shapley described and
formally analyzed a generalization of the Boston Pool algorithm and proved that
it computes a stable matching. Gale and Shapley used the metaphor of college
admissions. Essentially the same algorithm was independently developed by
Elliott Peranson in 1972 for use in medical school admissions. Similar algorithms
have since been adopted for many other matching markets, including faculty
hiring in France, hiring of new economics PhDs in the United States, university
admission in Germany, public school admission in New York and Boston, billet
assignments for US Navy sailors, and kidney-matching programs.

“This example was discovered by Donald Knuth.

170

4 - 5 - Stable Matching

Shapley was awarded the 2012 Nobel Prize in Economics for his research on
stable matchings, together with Alvin Roth, who significantly extended Shapley’s
work and used it to develop several real-world exchanges. (Gale did not share
the prize, because he died in 2008.)

Like our last failed greedy algorithm, the Gale-Shapley algorithm proceeds
in rounds until every position has been accepted. Each round has two stages:

r. An arbitrary unmatched hospital A offers its position to the best doctor a
(according to A’s preference list) who has not already rejected it.

2. If a is unmatched, she (tentatively) accepts A’s offer. If a already has
a match but prefers A, she rejects her current match and (tentatively)
accepts the new offer from A. Otherwise, a rejects the new offer.

Each doctor ultimately accepts the best offer that she receives, according to her
preference list. 11 In short, hospitals make offers greedily, and doctors accept
offers greedily. The doctors’ ability to reject their current matches in favor of
better offers is the key to making this mutual greedy strategy work.

For example, suppose that there are four doctors (Dr. Quincy, Dr. Rotwang,
Dr. Shephard, and Dr. Tam) and four hospitals (Arkham Asylum, Bethlem Royal
Hospital, County General Hospital, and The Dharma Initiative), who rank each
other as follows:

q r s t
A A B D
B D A B
C C C C
D B D A

A B C D
t r t s
s t r r
r q s q
q s q t

Given these preference lists as input, the Gale-Shapley algorithm might proceed
as follows:

r. Arkham makes an offer to Dr. Tam.

2. Bedlam makes an offer to Dr. Rotwang.

3. County makes an offer to Dr. Tam, who rejects her earlier offer from
Arkham.

4. Dharma makes an offer to Dr. Shephard. (From this point on, there is
only one unmatched hospital, so the algorithm has no more choices.)

5. Arkham makes an offer to Dr. Shephard, who rejects her earlier offer
from Dharma.

u The 1952 Boston Pool algorithm is a special case of the Gale-Shapley algorithm that executes
offers in a particular order. Roughly speaking, each offer is made by a hospital X whose favorite
doctor (among those who haven’t rejected X already) ranks X highest. Because the order of
offers depends on the entire set of preference lists, this algorithm must be executed by a central
authority; in contrast, the Gale-Shapley algorithm does not even require each participant to know
their own preferences in advance, as long as they behave consistently with some fixed rankings.

171

4 . Greedy Algorithms

6. Dharma makes an offer to Dr. Rotwang, who rejects her earlier offer from
Bedlam.

7. Bedlam makes an offer to Dr. Tam, who rejects her earlier offer from
County.

8. County makes an offer to Dr. Rotwang, who rejects it.

9. County makes an offer to Dr. Shephard, who rejects it.

10. County makes an offer to Dr. Quincy.

After the tenth round, all pending offers are accepted, and the algorithm returns
the matching {As, B t, Cq,Dr}. You can (and should) verify by brute force that
this matching is stable, even though no doctor was hired by her favorite hospital,
and no hospital hired their favorite doctor; in fact, County ended up hiring their
least favorite doctor. This is not the only stable matching for these preference
lists; the matching {Ar,Bs,Cq,Dt} is also stable.

Running Time

Analyzing the number of offers performed by the algorithm is relatively straight¬
forward (which is why we’re doing it first). Each hospital makes an offer to
each doctor at most once, so the algorithm makes at most n 2 offers.

To analyze the actual running time, however, we need to specify the algorithm
in more detail. How are the preference lists given to the algorithm? How does
the algorithm decide whether any hospital is unmatched, and if so, how does
it find an unmatched hospital? How does the algorithm store the tentative
matchings? How does the algorithm decide whether a doctor prefers to new
offer to her current match? Most fundamentally: How does the algorithm
actually represent doctors and hospitals?

One possibility is to represent each doctor and hospital by a unique integer
between 1 and n, and to represent preferences as two arrays Dpref[ 1 .. n, 1 .. n]
and Hpref[ 1 .. n, 1 .. n], where Dpref[i, r] represents the rth hospital in doctor
i’s preference list, and HPref[j,r] represents the rth doctor in hospital j’s
preference list. With the input in this form, the Boston Pool algorithm can
execute each offer in constant time, after some initial preprocessing; the overall
implementation runs in 0 (n 2 ) time. We leave the remaining details as a
straightforward exercise.

A somewhat harder exercise is to prove that there are inputs (and choices of
who makes offers when) that force Q(n 2 ) offers to be made before the algorithm
halts. Thus, our 0 (n 2 ) upper bound on the worst-case running time is tight.

172

4 - 5 - Stable Matching

Correctness

But why is the algorithm correct at all? How do we know that it always computes
a stable matching, or any complete matching for that matter? Gale and Shapley
argued correctness as follows.

The algorithm continues as long as there is at least one unfilled position;
conversely, when the algorithm terminates (after at most n 2 rounds), every
position is filled. No doctor can accept more than one position, and no hospital
can hire more than one doctor. So the algorithm always computes a matching.
(Whew!) It remains only to prove that the resulting matching is stable.

Suppose the algorithm matches some doctor a to some hospital A, even
though she prefers another hospital B. Because every doctor accepts the best
offer she receives, a received no offer she liked more than A; in particular, B
never made an offer to a. On the other hand, B made offers to every doctor
they prefer over their final match / 3 . It follows that B prefers [5 over a, which
means (a,B) is not an unstable pair. We conclude that there are no unstable
pairs; the matching is stable!

Optimality!

Surprisingly, the correctness of the Gale-Shapley algorithm does not depend
on which hospital makes its offer in each round. In fact, no matter which
unassigned hospital makes an offer in each round, the algorithm always computes
the same matching\ Let’s say that a is a feasible doctor for A if there is a stable
matching that assigns doctor a to hospital A.

Lemma 4.7. During the Gale-Shapley algorithm, each hospital A is rejected
only by doctors that are infeasible for A.

Proof: We prove the lemma by induction on the number of rounds. Consider
an arbitrary round of the algorithm, in which doctor a rejects one hospital A for
another hospital B. The rejection implies that a prefers B to A. Every doctor
that appears higher than a in B’s preference list was already rejected B in an
earlier round and therefore, by the inductive hypothesis, is infeasible for B.

Now consider an arbitrary matching (of the same doctors and hospitals)
that assigns a to A. We already established that a prefers B to A. If B prefers a
to its partner, the matching is unstable. On the other hand, if B prefers its
partner to a, then (by our earlier argument) its partner is infeasible, and again
the matching is unstable. We conclude that there is no stable matching that
assigns a to A. □

Now let best(A ) denote the highest-ranked feasible doctor on A’s preference
list. Lemma 4.7 implies that every doctor that A prefers to its final match is

173

4 . Greedy Algorithms

infeasible for A. On the other hand, the final matching is stable, so the doctor
assigned to A must be feasible for A. The following result is now immediate:

Corollary 4.8. The Gale-Shapley algorithm matches best(A ) with A, for every
hospital A.

In other words, the Gale-Shapley algorithm computes the best possible stable
matching from the hospitals’ point of view. It turns out that this matching is
also the worst possible from the doctors’ point of view! Let worsted ) denote the
lowest-ranked feasible hospital on doctor a’s preference list.

Corollary 4.9. The Gale-Shapley algorithm matches a with worst (a), for every
doctor a.

Proof: Suppose Gale and Shapley assign doctor a to hospital A; we need to
show that A = worst(a). Consider an arbitrary stable matching where A is not
matched with a but with another doctor / 3 . The previous corollary implies that
A prefers a = best(A ) to / 3 . Because the matching is stable, a must therefore
prefer her assigned hospital to A. This argument works for any stable matching,
so a prefers every other feasible match to A; in other words, A = worst(a). □

A subtle consequence of these two corollaries, discovered by Lester Dubins
and David Freedman in 1981, is that a doctor can potentially improve her match
by lying about her preferences, but a hospital cannot. (However, a set of hospitals
can collude so that some of their matches improve.) Partly for this reason, the
National Residency Matching Program reversed its matching algorithm in
1998, so that potential residents offer to work for hospitals, according to their
preference orders, and each hospital accepts its best offer. Thus, the new
algorithm computes the best possible stable matching for the doctors, and the
worst possible stable matching for the hospitals. In practice, however, this
reversal altered less than 1% of the residents’ matches. As far as I know, the
precise effect of this change on the patients is an open problem.

Exercises

Caveat lector: Some of these exercises cannot be solved using greedy
algorithms! Whenever you describe and analyze a greedy algorithm, you must
also include a proof that your algorithm is correct; this proof will typically take
the form of an exchange argument. These proofs are especially important in
classes (like mine) that do not normally require proofs of correctness.

174

Exercises

1. The GreedySchedule algorithm we described for the class scheduling
problem is not the only greedy strategy we could have tried. For each of
the following alternative greedy strategies, either prove that the resulting
algorithm always constructs an optimal schedule, or describe a small input
example for which the algorithm does not produce an optimal schedule.
Assume that all algorithms break ties arbitrarily (that is, in a manner that
is completely out of your control). [Hint: Three of these algorithms are
actually correct.]

(a) Choose the course x that ends last, discard classes that conflict with x,
and recurse.

(b) Choose the course x that starts first, discard all classes that conflict
with x, and recurse.

(c) Choose the course x that starts last, discard all classes that conflict
with x, and recurse.

(d) Choose the course x with shortest duration, discard all classes that
conflict with x, and recurse.

(e) Choose a course x that conflicts with the fewest other courses, discard all
classes that conflict with x, and recurse.

(f) If no classes conflict, choose them all. Otherwise, discard the course
with longest duration and recurse.

(g) If no classes conflict, choose them all. Otherwise, discard a course that
conflicts with the most other courses and recurse.

(h) Let x be the class with the earliest start time, and let y be the class with
the second earliest start time.

• If x and y are disjoint, choose x and recurse on everything but x.

• If x completely contains y, discard x and recurse.

• Otherwise, discard y and recurse.

(i) If any course x completely contains another course, discard x and
recurse. Otherwise, choose the course y that ends last, discard all classes
that conflict with y, and recurse.

2. Now consider a weighted version of the class scheduling problem, where
different classes offer different number of credit hours (totally unrelated
to the duration of the class lectures). Your goal is now to choose a set of
non-conflicting classes that give you the largest possible number of credit
hours, given arrays of start times, end times, and credit hours as input.

(a) Prove that the greedy algorithm described at the beginning of this
chapter—Choose the class that ends first and recurse—does not always
return an optimal schedule.

175

4 . Greedy Algorithms

(b) Prove that none of the greedy algorithms described in Exercise 1 always
return an optimal schedule. [Hint: Solve Exercise 1 first; the algorithms
that don’t work there don’t work here, either.]

(c) Describe and analyze an algorithm that always computes an optimal
schedule. [Hint: Your algorithm will not be greedy.]

3. LetX be a set of n intervals on the real line. We say that a subset of intervals
F cx covers X if the union of all intervals in Y is equal to the union of all
intervals in X. The size of a cover is just the number of intervals.

Describe and analyze an efficient algorithm to compute the smallest
cover of X. Assume that your input consists of two arrays L[ 1 .. n] and
i?[l.. zt], representing the left and right endpoints of the intervals in A. If
you use a greedy algorithm, you must prove that it is correct.

1 I i I I I □

I I I I I I

A set of intervals, with a cover (shaded) of size 7.

4. Let X be a set of n intervals on the real line. We say that a set P of points
stabs X if every interval in X contains at least one point in P. Describe and
analyze an efficient algorithm to compute the smallest set of points that
stabs X. Assume that your input consists of two arrays L [1 .. n] and R [1 .. n],
representing the left and right endpoints of the intervals in X. As usual. If
you use a greedy algorithm, you must prove that it is correct.

1 1 3

1 I 4

J

□

3 1

1 3

Zl [

4 l

ZD L

3

1

1

4

1

A set of intervals stabbed by four points (shown here as vertical segments)

5. LetX be a set of n intervals on the real line. A proper coloring oi X assigns a
color to each interval, so that any two overlapping intervals are assigned
different colors. Describe and analyze an efficient algorithm to compute the
minimum number of colors needed to properly color X. Assume that your
input consists of two arrays L[ 1 .. n] and R[ 1 .. n], representing the left and
right endpoints of the intervals in A. As usual, if you use a greedy algorithm,
you must prove that it is correct.

176

Exercises

1 I I 2 I I 3 1 I 4 |

A proper coloring of a set of intervals using five colors.

6 . (a) For every integer n, find a frequency array /[I.. n] whose Huffman
code tree has depth n — 1, such that the largest frequency is as small as
possible.

(b) Suppose the total length N of the unencoded message is bounded by a
polynomial in the alphabet size n. Prove that the any Huffman tree for
the frequencies /[I.. n] has depth O(logn).

r 7. Call a frequency array /[I.. n] a-heavy if it satisfies two conditions:

• /[l] > f[i] for all i > 1; that is, 1 is the unique most frequent symbol.

• /[l] > that is, at least an a fraction of the symbols are Is.

Find the largest real number a such that in every Huffman code for every
a-heavy frequency array, symbol 1 is represented by a single bit. [Hint:
First prove that 1/3 < a < 1 / 2 .J

8 . Describe and analyze an algorithm to compute an optimal ternary prefix-free
code for a given array of frequencies / [ 1 .. n ]. Don’t forget to prove that
your algorithm is correct for all n.

g. Describe in detail how to implement the Gale-Shapley stable matching
algorithm, so that the worst-case running time is 0(n 2 ), as claimed earlier
in this chapter.

to. (a) Prove that it is possible for the Gale-Shapley algorithm to perform Fi(n 2 )
offers before termination. (You need to describe both a suitable input
and a sequence of Q(n 2 ) valid offers.)

(b) Describe for any integer n a set of preferences for n doctors and n
hospitals that forces the Gale-Shapley algorithm to execute Fl(n 2 ) rounds,
no matter which valid proposal is made in each round. [Hint: Part (b)
implies part (a).]

n. Describe and analyze an efficient algorithm to determine whether a given
set of hospital and doctor preferences has to a unique stable matching.

12. Consider a generalization of the stable matching problem, where some
doctors do not rank all hospitals and some hospitals do not rank all doctors,

177

4. Greedy Algorithms

and a doctor can be assigned to a hospital only if each appears in the other’s
preference list. In this case, there are three additional unstable situations:

• A matched hospital prefers an unmatched doctor to its assigned match.

• A matched doctor prefers an unmatched hospital to her assigned match.

• An unmatched doctor and an unmatched hospital appear in each other’s

preference lists.

A stable matching in this setting may leave some doctors and/or hospitals
unmatched, even though their preference lists are non-empty. For example,
if every doctor lists Harvard as their only acceptable hospital, and every
hospital lists Dr. House as their only acceptable intern, then only House and
Harvard will be matched.

Describe and analyze an efficient algorithm that computes a stable
matching in this more general setting. [Hint: Reduce to an instance
where every doctor ranks every hospital and vice versa, and then invoke
Gale-Shapley.]

13. The Scandinavian furniture company Ftirni has hired n drivers to deliver
n identical orders to n different addresses in Wilmington, Delaware. Each
driver has their own well-established delivery route through Wilmington
that visits all n addresses. Assuming they follow their routes as they always
do, two drivers never visit the same addresses at the same time.

In principle, each of the n drivers can deliver their furniture to any of
the n addresses, but there’s a complication. One of the drivers has secretly
wired proximity sensors and explosives to the Johannshamn sofas (with the
Strinne green stripe pattern). If two sofas are ever at the same address at
the same time, both will explode, destroying both the delivery truck and
the building at that address. This can only happen if one driver delivers an
order to that address, and then later another driver visits that same address
while the furniture is still on their truck.

Your job as the Fiirni dispatcher is to assign each driver to a delivery
address. Describe an algorithm to assign addresses to drivers so that each of
the n addresses receives their furniture order and there are no explosions.

For example, suppose Jack’s route visits 537 Paper Street at 6pm and
1888 Franklin Street at 8pm, and Marla’s route visits 537 Paper at 7pm and
1888 Franklin at 9pm. Then Jack should deliver to 1888 Franklin, and Marla
should deliver to 537 Paper; otherwise, there would be an explosion at 1888
Franklin at 8pm. (Cue the Pixies.) [Hint: Jack and Marla are a bit unstable.]

14. Suppose you are a simple shopkeeper living in a country with n different
types of coins, with values 1 = c[l] < c[ 2 ] < ••• < c[n], (In the U.S.,

178

Exercises

for example, n = 6 and the values are 1 , 5 , 10 , 25 , 50 and 100 cents.)
Your beloved and benevolent dictator, El Generalissimo, has decreed that
whenever you give a customer change, you must use the smallest possible
number of coins, so as not to wear out the image of El Generalissimo lovingly
engraved on each coin by servants of the Royal Treasury.

(a) In the United States, there is a simple greedy algorithm that always
results in the smallest number of coins: subtract the largest coin and
recursively give change for the remainder. El Generalissimo does not
approve of American capitalist greed. Show that there is a set of coin
values for which the greedy algorithm does not always give the smallest
possible of coins.

(b) Now suppose El Generalissimo decides to impose a currency system
where the coin denominations are consecutive powers b°, b 1 , b 2 ,..., b k
of some integer b > 2 . Prove that despite El Generalissimo’s disapproval,
the greedy algorithm described in part (a) does make optimal change in
this currency system.

(c) Describe and analyze an efficient algorithm to determine, given a target
amount T and a sorted array c[l.. n] of coin denominations, the smallest
number of coins needed to make T cents in change. Assume that
c[l] = 1 , so that it is possible to make change for any amount T.

15. Suppose you are given an array A[ 1 .. n] of integers, each of which may be
positive, negative, or zero. A contiguous subarray A[i ..;] is called a positive
internal if the sum of its entries is greater than zero. Describe and analyze
an algorithm to compute the minimum number of positive intervals that
cover every positive entry in A. For example, given the following array as
input, your algorithm should output 3 . If every entry in the input array is
negative, your algorithm should output 0.

sum=2 sum=l sum=7

+3

-5

+7

-4

+1

-8

+3

-7

+5

-9

+5

-2

+4

16. Consider the following process. At all times you have a single positive
integer x, which is initially equal to 1 . In each step, you can either
increment x or double x. Your goal is to produce a target value n. For
example, you can produce the integer 10 in four steps as follows:

Obviously you can produce any integer n using exactly n — 1 increments, but
for almost all values of n, this is horribly inefficient. Describe and analyze
an algorithm to compute the minimum number of steps required to produce
any given integer n.

179

4. Greedy Algorithms

17. Suppose we have n skiers with heights given in an array P[ 1 .. n], and n skis
with heights given in an array S[ 1 .. n\. Describe an efficient algorithm to
assign a ski to each skier, so that the average difference between the height
of a skier and her assigned ski is as small as possible. The algorithm should
compute a permutation cr such that the expression

is as small as possible.

18. Alice wants to throw a party and she is trying to decide who to invite. She
has n people to choose from, and she knows which pairs of these people
know each other. She wants to pick as many people as possible, subject to
two constraints:

• For each guest, there should be at least five other guests that they already

know.

• For each guest, there should be at least five other guests that they don’t
already know.

Describe and analyze an algorithm that computes the largest possible number
of guests Alice can invite, given a list of n people and the list of pairs who
know each other.

19. Suppose we are given two arrays C[ 1 .. n] and R[ 1 .. n] of positive integers.
An n x n matrix of Os and Is agrees with R and C if, for every index i, the
ith row contains R[i] Is, and the ith column contains C[i] Is. Describe
and analyze an algorithm that either constructs a matrix that agrees with R
and C, or correctly reports that no such matrix exists.

20. You’ve just accepted a job from Elon Musk, delivering burritos from San
Francisco to New York City. You get to drive a Burrito-Delivery Vehicle
through Elon’s new Transcontinental Underground Burrito-Delivery Tube,
which runs in a direct line between these two cities.

Your Burrito-Delivery Vehicle runs on single-use batteries, which must
be replaced after at most 100 miles. The actual fuel is virtually free, but
the batteries are expensive and fragile, and therefore must be installed only
by official members of the Transcontinental Underground Burrito-Delivery
Vehicle Battery-Replacement Technicians’ Union. 12 Thus, even if you replace
your battery early, you must still pay full price for each new battery to be

12 or as they call themselves in German, Die Transkontinentaluntergrundburritolieferfahrzeug-
batteriewechseltechnikervereinigung.

180

Exercises

installed. Moreover, your Vehicle is too small to carry more than one battery
at a time.

There are several fueling stations along the Tube; each station charges a
different price for installing a new battery. Before you start your trip, you
carefully print the Wikipedia page listing the locations and prices of every
fueling station along the Tube. Given this information, how do you decide
the best places to stop for fuel?

More formally, suppose you are given two arrays D[ 1 .. n] and C [1 .. n],
where D[i] is the distance from the start of the Tube to the ith station, and
C[i] is the cost to replace your battery at the ith station. Assume that your
trip starts and ends at fueling stations (so D[ 1 ] = 0 and D[n] is the total
length of your trip), and that your car starts with an empty battery (so you
must install a new battery at station 1).

(a) Describe and analyze a greedy algorithm to find the minimum number
of refueling stops needed to complete your trip. Don’t forget to prove
that your algorithm is correct.

(b) But what you really want to minimize is the total cost of travel. Show
that your greedy algorithm in part (a) does not produce an optimal
solution when extended to this setting.

(c) Describe an efficient algorithm to compute the locations of the fuel
stations you should stop at to minimize the total cost of travel.

2i. You’ve been hired to store a sequence of n books on shelves in a library. The
order of the books is fixed by the cataloging system and cannot be changed;
each shelf must store a contiguous interval of the given sequence of books.
You are given two arrays H [1 .. n] and T [1 .. n], where H[i] and T[i] are
respectively the height and thickness of the ith book in the sequence. All
shelves in this library have the same length L; the total thickness of all books
on any single shelf cannot exceed L.

(a) Suppose all the books have the same height h and the shelves have height
larger than h, so every book fits on every shelf. Describe and analyze a
greedy algorithm to store the books in as few shelves as possible. [Hint:
The algorithm is obvious, but why is it correct?]

(b) That was a nice warmup, but now here’s the real problem. In fact the
books have different heights, but you can adjust the height of each shelf
to match the tallest book on that shelf. (In particular, you can change
the height of any empty shelf to zero.) Now your task is to store the
books so that the sum of the heights of the shelves is as small as possible.
Show that your greedy algorithm from part (a) does not always give the
best solution to this problem.

181

4. Greedy Algorithms

(c) Describe and analyze an algorithm to find the best matching between
books and shelves as described in part (b).

22. A string w of parentheses ( and ) is balanced if it satisfies one of the
following conditions:

• w is the empty string.

• w = (x) for some balanced string x

• w = xy for some balanced strings x and y
For example, the string

w = ((())()()) (()()) ()

is balanced, because w = xy, where

*=((0)00) and y=(()())().

(a) Describe and analyze an algorithm to determine whether a given string
of parentheses is balanced.

(b) Describe and analyze a greedy algorithm to compute the length of a
longest balanced subsequence of a given string of parentheses. As usual,
don’t forget to prove your algorithm is correct.

For both problems, your input is an array w[l.. n], where for each i, either
w[i] = ( or w[i] = ). Both of your algorithms should run in 0 (n) time.

23. One day Alex got tired of climbing in a gym and decided to take a large
group of climber friends outside to climb. They went to a climbing area
with a huge wide boulder, not very tall, with several marked hand and foot
holds. Alex quickly determined an “allowed” set of moves that her group of
friends can perform to get from one hold to another.

The overall system of holds can be described by a rooted tree T with n
vertices, where each vertex corresponds to a hold and each edge corresponds
to an allowed move between holds. The climbing paths converge as they go
up the boulder, leading to a unique hold at the summit, represented by the
root of T.

Alex and her friends (who are all excellent climbers) decided to play a
game, where as many climbers as possible are simultaneously on the boulder
and each climber needs to perform a sequence of exactly k moves. Each
climber can choose an arbitrary hold to start from, and all moves must move
away from the ground. Thus, each climber traces out a path of k edges
in the tree T, all directed toward the root. However, no two climbers are
allowed to touch the same hold; the paths followed by different climbers
cannot intersect at all.

182

Exercises

(a) Describe and analyze a greedy algorithm to compute the maximum
number of climbers that can play this game. Your algorithm is given
a rooted tree T and an integer k as input, and it should compute the
largest possible number of disjoint paths in T, where each path has
length k. Do not assume that T is a binary tree. For example, given the
tree below as input, your algorithm should return the integer 8.

Figure 4.5. Seven disjoint paths of length k = 3. This is not the largest such set of paths in this tree.

(b) Now suppose each vertex in T has an associated reward, and your goal
is to maximize the total reward of the vertices in your paths, instead of
the total number of paths. Show that your greedy algorithm does not
always return the optimal reward.

(c) Describe an efficient algorithm to compute the maximum possible reward,
as described in part (b).

24. Congratulations! You have successfully conquered Camelot, transforming
the former battle-scarred hereditary monarchy into an anarcho-syndicalist
commune, where citizens take turns to act as a sort of executive-officer-
for-the-week, but with all the decisions of that officer ratified at a special
bi-weekly meeting, by a simple majority in the case of purely internal affairs,
but by a two-thirds majority in the case of more major....

As a final symbolic act, you order the Round Table (surprisingly, an
actual circular table) to be split into pizza-like wedges and distributed to the
citizens of Camelot as trophies. Each citizen has submitted a request for an
angular wedge of the table, specified by two angles—for example: Sir Robin
the Brave might request the wedge from 17.23° to 42°, and Sir Lancelot
the Pure might request the 2° wedge from 359° to 1°. Each citizen will be
happy if and only if they receive precisely the wedge that they requested.
Unfortunately, some of these ranges overlap, so satisfying all the citizens’
requests is simply impossible. Welcome to politics.

183

4. Greedy Algorithms

Describe and analyze an algorithm to find the maximum number of
requests that can be satisfied. [Hint: The output of your algorithm should
not change if you rotate the table. Do not assume that angles are integers.]

25 . Suppose you are standing in a field surrounded by several large balloons.
You want to use your brand new Acme Brand Zap-0-Matic™ to pop all the
balloons, without moving from your current location. The Zap-O-Matic™
shoots a high-powered laser beam, which pops all the balloons it hits. Since
each shot requires enough energy to power a small country for a year, you
want to fire as few shots as possible.

Figure 4.6. Nine balloons popped by four shots of the Zap-O-Matic™

The minimum zap problem can be stated more formally as follows. Given
a set C of n circles in the plane, each specified by its radius and the (x,y)
coordinates of its center, compute the minimum number of rays from the
origin that intersect every circle in C. Your goal is to find an efficient
algorithm for this problem.

(a) Suppose it is possible to shoot a ray that does not intersect any balloons.
Describe and analyze a greedy algorithm that solves the minimum zap
problem in this special case. [Hint: See Exercise 2 .]

(b) Describe and analyze a greedy algorithm whose output is within 1 of
optimal. That is, if m is the minimum number of rays required to hit
every balloon, then your greedy algorithm must output either morm + 1 .
(Of course, you must prove this fact.)

(c) Describe an algorithm that solves the minimum zap problem in 0(n 2 )
time.

r (d) Describe an algorithm that solves the minimum zap problem in 0(n log n )
time.

Assume you have a subroutine that tells you the range of angles of rays that
intersects an arbitrary circle c in 0(1) time. This subroutine is not difficult
to write, but it’s not the interesting part of the problem.

184

[T]he distributions and partitions of knowledge are not like several lines that meet
in one angle, and so touch but in a point, but are like branches of a tree that meet in
a stem, which hath a dimension and quantity of entireness and continuance before
it come to discontinue and break itself into arms and boughs.

— Francis Bacon, The Advancement of Learning (1605)

Thus you see, most noble Sir, how this type of solution bears little relationship to
mathematics, and I do not understand why you expect a mathematician to
produce it, rather than anyone else.

— Leonhard Euler, describing the Konigsburg bridge problem
in a letter to Carl Leonhard Gottlieb Ehler (April 3,1736)

Well, ya turn left by the fire station in the village and take the old post road by the
reservoir and... no, that won’t do.

Best to continue straight on by the tar road until you reach the schoolhouse and
then turn left on the road to Bennett's Lake until... no, that won't work either.

East Millinocket, ya say? Come to think of it, you can’t get there from here.

— Robert Bryan and Marshall Dodge,
Bert and I and Other Stories from Down East (1961)

5

Basic Graph Algorithms

5.1 Introduction and History

A graph is a collection of pairs—pairs of integers, pairs of people, pairs of
cities, pairs of stars, pairs of countries, pairs of scientific papers, pairs of web
pages, pairs of game positions, pairs of recursive subproblems, even pairs
of graphs. Mirroring the most common method for visualizing graphs, the
underlying objects being paired are usually called vertices or nodes, and the
pairs themselves are called edges or arcs, but in fact the objects and pairs can
be anything at all.

One of the earliest examples of graphs are road networks and maps thereof.
Roman engineers constructed a network of more than 400 000 km of public
roads across Europe, western and central Asia, and northern Africa during
the height of the Roman empire. Travelers on the road network would carry
itineraria, which were either simple lists or more pictorial representations of
the landmarks and distances along various roads. The Tabula Peutingeriana, a

185

5. Basic Graph Algorithms

13th-century scroll depicting the entire Roman cursus publicus, is widely believed
to be a medieval copy of a 5th-century revision of a ist-century itinerarium
pictum, commissioned during the reign of Augustus Caesar. The Peutinger Table
is not a geographically accurate map—historians debate whether it qualifies as
a “map” at all!—but an abstract representation of the road network, similar to
a modern subway map. Cities along each road are indicated by kinks in the
curve representing that road; the names of these cites and the lengths of road
segments between them are also indicated on the map. Thus, the map contains
enough information to find the shortest route between any two cities in the
5th-century Roman empire. See Figure 5.1.

Figure 5.1. A small excerpt of Konrad Miller's 1872 restoration of the Tabula Peutingerlana, showing the
Roman road from modern-day Birten ( Veteribus , top left) through Koln (, Agripina ) and Bonn ( Bonnae ) to
Mainz ( Mogontiaco, top right), with branches to Trier ( Avg Tresvirorvm, center) and Metz ( Matrlcorvm,
bottom center). (See Image Credits at the end of the book.)

One of the oldest classical applications of graphs—and specifically trees—is
in representing genealogies. Complex family “trees” have been used for centuries
to settle legal questions about marriage, inheritance, and royal succession. Civil
law in the Roman empire, later adopted as canon law by the early Catholic
Church, forbade marriage between first cousins or closer relatives. In the early
ninth century, the Church changed both the required distance and the method
of computation. Where the Roman computatio legalis required the sum of
the distances to the nearest common ancestor to be at least four, the newer
computatio canonica required the maximum of the two distances to be at least
seven. In 1215, bowing to practical considerations (and actual practice), the
Church relaxed the minimum required distance for marriage to four. 1 The
left diagram in Figure 5.2 illustrates a particularly convoluted case: Tirius and
Theburga marry and have a son Gaius, after which Tirius dies; Theburga then

‘During the 11th and 12th centuries, this restriction gradually expanded to include up to four
links by affinity, initially through marriage, and later through extra-marital sex, betrothal, and
even godparenting. For example, marriage between a man and his sister’s husband’s sister’s
husband’s sister was formally forbidden, as was a marriage between a widower and his son’s
wife’s widowed mother. These affinity requirements were significantly reduced but not eliminated
in 1215; the Church only abandoned the concept of affinity ex copula illicita in 1917.

186

5 - 1 . Introduction and History

marries Lothar, bears him a son, and dies; finally, Lothar and Bertha marry and
have a daughter Gemma. Can Gaius’s son legally marry Gemma’s daughter?

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Figure 5.2. Two diagrams describing a complex marriage case, from an anonymous 15th-century trea¬
tise on Johannes Andreae's Super arboribus consanguinitatis et affmitatis, an early 14th-century treatise
on canon law. (See Image Credits at the end of the book.)

In the mid-i68os, French mathematician Pierre Varignon developed a graph¬
ical method for finding the equilibrium position of a tree-like network of ropes
under tension, building on earlier work by Simon Stevin published a century
earlier. Varignon observed that when the ropes are at equilibrium, one can draw
a graph whose edges are segments parallel to the ropes, with lengths equal to
the forces along those ropes, such that the ropes meeting at any point in the
network define a closed cycle in the graph. Varignon’s graphical method was not
published in complete detail until 1725, two years after his death. These graphs
are now known as reciprocal force diagrams or Maxwell-Cremona diagrams.

Figure 5.3. Reciprocal force diagrams (dotted), from Varignon's posthumous Nouvelle mecanique, ou
statique, dont le projet fut donne en MDCLXXXVII [New mechanics, or statics, whose project was given
in t68y] (See Image Credits at the end of the book.)

Of course, there are many other familiar examples of graphs, like board
games (dating to antiquity); vertices and edges of convex polyhedra (formally
studied by ancient Greek philosophers, but much older); visualizations of

187

5. Basic Graph Algorithms

star patterns (already developed in East Asia by the 7th century ce); mazes
(introduced in their modern form by Giovanni Fontana circa 1420); geodetic
triangulations (introduced by Gemma Frisius in 1533, and used to calculate the
circumference of the earth by Willebrod Snell in 1615 and to define the meter in
1799), Leonard Euler’s well-known partial 2 solution to the Bridges of Konigsburg
puzzle (1735) or his less well-known complete solution of the knight’s tour
problem (1759); telegraph and other communication networks (first proposed
in 1753, developed by Ronalds, Schilling, Gauss, Weber, and others in the early
1800s, and deployed worldwide by the late 1800s); electrical circuits (formalized
in the early 1800s by Ohm, Maxwell, Kirchoff, and others); molecular structural
formulas (introduced independently by August Kekule in 1857 and Archibald
Couper in 1858); social networks (first studied in the mid-i930s by sociologist
Jacob Moreno); digital electronic circuits (proposed by Charles Sanders Pierce
in 1886, and cast into their modern form by Claude Shannon in 1937); and yeah,
okay, if you insist, the modern internet.

The word “graph” for the abstract mathematical was coined by James
Sylvester in 1878, who adapted Kekule’s “chemicographs” to describe certain
algebraic invariants, at the suggestion of his colleague William Clifford. The
word “tree” was first used for connected acyclic graphs by Arthur Cayley in 1857,
although the abstract concept of trees had already been used by Gustav Kirchoff
and Karl von Staudt ten years earlier. The zeroth book on graph theory was
published by Andre Sainte-Lague in 1926; Denes Konig published the first graph
theory book ten years later.

5.2 Basic Definitions

Formally, a (simple) graph is a pair of sets (V,F), where V is an arbitrary
non-empty finite set, whose elements are called vertices 3 or nodes, and E is a
set of pairs of elements of V, which we call edges. In an undirected graph, the
edges are unordered pairs, or just sets of size two; I usually write uv instead
of {u,v} to denote the undirected edge between u and v. In a directed graph,
the edges are ordered pairs of vertices; I usually write u->v instead of (u, v) to
denote the directed edge from u to v.

2 Euler dismissed the final step of his argument—actually finding an Euler tour of a graph
when every vertex has even degree—as obvious. Euler also failed to notice that a graph with an
Euler tour must be connected. The first complete proof that a graph has an Euler tour if and only
if it is connected and every vertex has even degree was published by Carl Hierholzer in 1873.

3 The singular of the English word “vertices” is vertex. Similarly, the singular of “matrices” is
matrix, and the singular of “indices” is index. Unless you’re speaking Italian, there is no such
thing as a vertice, matrice, indice, appendice, helice, apice, vortice, radice, simplice, codice,
directrice, dominatrice, Unice, Kleenice, Asterice, Obelice, Dogmatice, Getafice, Cacofonice,
Vitalstatistice, Geriatrice, or Jimi Hendrice! If you have trouble remembering this rule, stick to

188

5.2. Basic Definitions

Following standard (but admittedly confusing) practice, I will also use V to
denote the number of vertices in a graph, and E to denote the number of edges.
Thus, in any undirected graph we have 0 < E < (2), and in any directed graph
we have 0 < E < V(V — 1 ).

The endpoints of an edge uv or u-*v are its vertices u and v. We distinguish
the endpoints of a directed edge u->v by calling u the tail and v the head.

The definition of a graph as a pair of sets forbids multiple undirected edges
with the same endpoints, or multiple directed edges with the same head and
the same tail. (The same directed graph can contain both a directed edge u->v
and its reversal v-»u.) Similarly, the definition of an undirected edge as a set
of vertices forbids an undirected edge from a vertex to itself. Graphs without
loops and parallel edges are often called simple graphs; non-simple graphs
are sometimes called multigraphs. Despite the formal definitional gap, most
algorithms for simple graphs extend to multigraphs with little or no modification,
and for that reason, I see no need for a formal definition here.

For any edge uv in an undirected graph, we call u a neighbor of v and vice
versa. The degree of a node is its number of neighbors. In directed graphs,
we distinguish two kinds of neighbors. For any directed edge u-»v, we call u a
predecessor or in-neighbor of v and v a successor or out-neighbor of u. The
in-degree of a node is the number of predecessors; the out-degree is the number
of successors.

A graph G' = (V 7 , E') is a subgraph of G = (V,E) if V 7 c y and E' c E.
A proper subgraph of G is any subgraph other than G itself.

A walk in an undirected graph is a sequence of vertices, where each
successive pair of vertices are adjacent; informally, we can also think of a walk as
a sequence of edges. A walk is called a path if it visits each vertex at most once.
For any two vertices u and v in a graph G, we say that v is reachable from u
if G contains a walk (and therefore a path) between u and v. An undirected
graph is connected if every vertex is reachable from every other vertex. Every
undirected graph consists of one or more components, which are its maximal
connected subgraphs; two vertices are in the same component if and only if
there is a path between them. 4

A cycle is a path that starts and ends at the same vertex and has at least one
edge. An undirected graph is acyclic if no subgraph is a cycle; acyclic graphs
are also called forests. A tree is a connected acyclic graph, or equivalently, one
component of a forest. A spanning tree of an undirected graph G is a subgraph
that is a tree and contains every vertex of G. A graph has a spanning tree if and
only if it is connected. A spanning forest of G is a collection of spanning trees,
one for each component of G.

“node”.

Components are often called “connected components”, but this usage is redundant; compo¬
nents are connected by definition.

189

5. Basic Graph Algorithms

Directed graphs require slightly different definitions. A directed walk is
a sequence of directed edges, where the head of each edge is the tail of the
next; a directed path is a directed walk without repeated vertices. Vertex v
is reachable from vertex u in a directed graph G if and only if G contains a
directed walk (and therefore a directed path) from u to v. A directed graph
is strongly connected if every vertex is reachable from every other vertex. A
directed graph is acyclic if it does not contain a directed cycle; directed acyclic
graphs are often called dags.

5.3 Representations and Examples

The most common way to visually represent graphs is by drawing them. A
drawing of a graph maps each vertex to a point in the plane (typically drawn as
a small circle or some other shape) and each edge to a curve or straight line
segment between the two vertices. A graph is planar if it has a drawing where
no two edges cross; such a drawing is also called an embedding . 5 The same
graph can have many different drawings, so it is important not to confuse a
particular drawing with the graph itself. In particular, planar graphs can have
non-planar drawings!

Figure 5.4. Two drawings of the same disconnected planar graph with 13 vertices, 19 edges, and two
components. Only the drawing on the left is an embedding.

However, drawings are far from the only useful representation of graphs.
For example, the intersection graph of a collection of geometric objects has a
node for every object and an edge for every intersecting pair of objects. Whether
a particular graph can be represented as an intersection graph depends on what
kind of object you want to use for the vertices. Different types of objects—line
segments, rectangles, circles, etc.—define different classes of graphs. One
particularly useful type of intersection graph is an interval graph, whose vertices
are intervals on the real line, with an edge between any two intervals that
overlap.

Another good example is the dependency graph of a recursive algorithm.
Dependency graphs are directed acyclic graphs. The vertices are all the distinct
recursive subproblems that arise when executing the algorithm on a particular

5 Confusingly, the word “embedding” is often used as a synonym for “drawing”, even when
the edges intersect. Please don’t do that.

190

5 - 3 - Representations and Examples

Figure 5.5. The graph in Figure 5.4 is also the intersection graph of (a) a set of line segments and (b) a
set of circles.

input. There is an edge from one subproblem to another if evaluating the second
subproblem requires a recursive evaluation of the first. For example, for the
Fibonacci recurrence

F

n

0 if n = 0,

< 1 if n = 1,

F n _ : + F n _ 2 otherwise,

the vertices of the dependency graph are the integers 0,1,2,...,n, and the
edges are the pairs (i — l)-u and (i — 2)->i for every integer i between 2 and n.

Figure 5.6. The dependency graph of the Piiigala-Fibonacci recurrence.

As a more complex example, recall the recurrence for the edit distance
problem from Chapter 3:

Edit(i,j)= <

J

mm <

Edit(i — 1 , j) + 1
Edit{i,j — 1 ) + 1
Edit(i-l,j-l) + [A[i]^B[j]],

if; = 0
if i = 0

otherwise

The dependency graph of this recurrence is an m x n grid of vertices (i,;)
connected by vertical edges (i — 1 horizontal edges (i,; — l)->(i,;),

and diagonal edges (i — 1 , j — 1 )—>(i,;). Dynamic programming works efficiently
for any recurrence that has a reasonably small dependency graph; a proper
evaluation order ensures that each subproblem is visited after its predecessors.

Another interesting example is the configuration graph of a game, puzzle,
or mechanism like tic-tac-toe, checkers, the Rubik’s Cube, the Tower of Hanoi,
or a Turing machine. The vertices of the configuration graph are all the valid
configurations of the puzzle; there is an edge from one configuration to another

5. Basic Graph Algorithms

Figure 5.7. The dependency graph of the edit distance recurrence.

if it is possible to transform one configuration into the other with a simple move.
(Obviously, the precise definition depends on what moves are allowed.) Even
for reasonably simple mechanisms, the configuration graph can be extremely
complex, and we typically only have access to local information about the
configuration graph.

Figure 5.8. The configuration graph of the 4-disk Tower of Hanoi.

Finite-state automata used in formal language theory can be modeled as
labeled directed graphs. Recall that a deterministic finite-state automaton is
formally defined as a 5-tuple M = (S, Q,s,A, 5 ), where £ is a finite set called
the alphabet, Q is a finite set of states, s e Q is the start state, A c Q is the set of
accepting states, and 5 :QxS—>Qisa transition function. But it is often more
useful to think of M as a directed graph G M whose vertices are the states Q,
and whose edges have the form q^ 5 (q,a) for every state q e Q and symbol
a G £. Basic questions about the language L(M) accepted by M can then be
phrased as questions about the graph G M . For example, L(M) = 0 if and only
if no accepting state/vertex is reachable from the start state/vertex s.

Finally, sometimes one graph can be used to implicitly represent other larger
graphs. A good example of this implicit representation is the subset construction,

192

54 - Data Structures

which is normally used to convert NFAs into DFAs, but can be applied to arbitrary
directed graphs as follows. Given any directed graph G = (V, E), we can define
a new directed graph G' = ( 2 y , E'j whose vertices are all subsets of vertices in V,
and whose edges E' are defined as follows:

E' := {A-^B | u->v € E for some u € A and v € B }

We can mechanically translate this definition into an algorithm to construct G'
from G, but strictly speaking, this construction is unnecessary, because G is

already an implicit representation of G'.

It’s important not to confuse any of these examples/representations with the
actual formal definition: A graph is a pair of sets (V,E), where V is an arbitrary
non-empty finite set, and £ is a set of pairs (either ordered or unordered) of
elements of V. In short: A graph is a set of pairs of things.

5.4 Data Structures

In practice, graphs are usually represented by one of two standard data struc¬
tures: adjacency lists and adjacency matrices. At a high level, both data structures
are arrays indexed by vertices; this requires that each vertex has a unique integer
identifier between 1 and V. In a formal sense, these integers are the vertices.

Adjacency Lists

By far the most common data structure for storing graphs is the adjacency list.
An adjacency list is an array of lists, each containing the neighbors of one of the
vertices (or the out-neighbors if the graph is directed). 6 For undirected graphs,
each edge uv is stored twice, once in u’s neighbor list and once in v’s neighbor
list; for directed graphs, each edge u->v is stored only once, in the neighbor
list of the tail u. For both types of graphs, the overall space required for an
adjacency list is 0 (V + E).

There are several different ways to represent these neighbor lists, but the
standard implementation uses a simple singly-linked list. The resulting data
structure allows us to list the (out-)neighbors of a node v in 0(1 + deg(v)) time;
just scan v’s neighbor list. Similarly, we can determine whether u->v is an edge
in 0(1 + deg(u)) time by scanning the neighbor list of u. For undirected graphs,
we can improve the time to 0(1 + min{deg(u), deg(v)}) by simultaneously
scanning the neighbor lists of both u and v, stopping either we locate the edge
or when we fall of the end of a list.

6 Attentive students might notice that despite is name, an adjacency list is not a list. This
nomenclature is an example of the Red Herring Principle: In computer science, as in mathematics,
a red herring is neither necessarily red nor necessarily a fish.

193

5. Basic Graph Algorithms

Figure 5.9. An adjacency list for our example graph.

Of course, linked lists are not the only data structure we could use; any
other structure that supports searching, listing, insertion, and deletion will do.
For example, we can reduce the time to determine whether uv is an edge to
0(1 + log(deg(u))) by using a balanced binary search tree to store the neighbors
of u, or even to 0(1) time by using an appropriately constructed hash table. 7

One common implementation of adjacency matrices is the adjacency array,
which uses a single array to store all edge records, with the records of edges
incident to each vertex in a contiguous interval, and with a separate array
storing the index of the first edge incident to each vertex. Moreover, it is useful
to keep the intervals for each vertex in sorted order, as shown in the figure
below, so that we can check in 0(logdeg(iz)) time whether two vertices u and v
are adjacent.

abed

e /

g h i j k l m

) J

1 1 1 1 -.

n 1 rP-

J

b I e I a I c | e |/| b \ d \f \ g | c | g |q | b |/ | g | h \ b | c] f | ,^ | c | d \ e \ f \ i \ e | g | fc | I |m| j \ l \ j \ k |m| j |T|

abcdefghijklm

1

3

7

11

13

18

22

27

28

29

32

34

37

|2|5|l|3|5|6|2|4|6|7|3|7|l|a|b|7|s|a|3|5|7|3|4|5|6|9|5|7 |ll 112113|l0 |l2|l0|ll |l3|lO |l2|

Figure 5.10. An abstract adjacency array for our example graph, and its actual implementation as a
pair of arrays.

Adjacency Matrices

The other standard data structure for graphs is the adjacency matrix , 8 first
proposed by George Brunei in 1894. The adjacency matrix of a graph G is a
VxV matrix of Os and Is, normally represented by a two-dimensional array

7 This is a lot more subtle than it sounds. Most popular hashing techniques do not guarantee
fast query times, and even most good hashing methods can guarantee only 0 ( 1 ) expected time.
See http://algorithms.wtf for a more thorough discussion of hashing.

8 See footnote 3 .

194

54 - Data Structures

A [1 ,.V ,1 ..V], where each entry indicates whether a particular edge is present
in G.

• If the graph is undirected, then A[u, v] := [izv € E] for all vertices u and v.

• If the graph is directed, then A[u, v] := [iz->v e E] for all vertices u and v.

For undirected graphs, the adjacency matrix is always symmetric, meaning
A[u,v ] = A[v, iz] for all vertices zi and v, because uv and viz are just different
names for the same edge, and the diagonal entries A[u, iz] are all zeros. For
directed graphs, the adjacency matrix may or may not be symmetric, and the
diagonal entries may or may not be zero.

a

b

c

d

e

f

g

h

i

i

k

l

m

abcdefghijklm

0 10 0 10
10 10 11
0 10 10 1
0 0 1 0 0 0
1 1 0 0 0 1
0 110 10
0 0 1111
0 0 0 0 1 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0

0 0 0 0 0 0 0
0 0 0 0 0 0 0
1 0 0 0 0 0 0
1 0 0 0 0 0 0
1 1 0 0 0 0 0
1 0 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
1 0 0 0 0 0 0
0 0 0 0 1 1 1
0 0 0 1 0 1 0
0 0 0 1 1 0 1
0 0 0 1 0 1 0

Figure 5.11. An adjacency matrix for our example graph.

Given an adjacency matrix, we can decide in 0 ( 1 ) time whether two vertices
are connected by an edge just by looking in the appropriate slot in the matrix.
We can also list all the neighbors of a vertex in 0 ( 1 /) time by scanning the
corresponding row (or column). This running time is optimal in the worst case,
but even if a vertex has few neighbors, we still have to scan the entire row to
find them all. Similarly, adjacency matrices require 0 (V 2 ) space, regardless of
how many edges the graph actually has, so they are only space-efficient for very
dense graphs.

Comparison

Table 5.1 summarizes the performance of the various standard graph data
structures. Stars* indicate expected amortized time bounds for maintaining
dynamic hash tables. 9

In light of this comparison, one might reasonably wonder why anyone would
ever use an adjacency matrix; after all, adjacency lists with hash tables support
the same operations in the same time, using less space. The main reason is that
for sufficiently dense graphs, adjacency matrices are simpler and more efficient

9 Don’t worry if you don’t understand the phrase “expected amortized”.

195

5. Basic Graph Algorithms

Standard adjacency list
(linked lists)

Fast adjacency list
(hash tables)

Adjacency

matrix

Space

0(V + £)

0(V + £)

0(V 2 )

Test if uv 6 E

Test if u-»v 6 £

List v's (out-)neighbors
List all edges

Insert edge uv

Delete edge uv

0(1 + min{deg(u), deg(v)}) = 0(V)
0(1 + deg(u)) = 0(V)

0(1 + deg(v)) = 0(V)

0(V + £)

0 (1)

0(deg(u) + deg(v)) = 0(V)

0 (1)

0 (1)

0(1 + deg(v)) = 0(V)
0(V + £)

0 (1)*

0 (1)*

0 (1)

0 (1)

0(V)

0(V 2 )

0 (1)

0 (1)

Table 5.1 . Times for basic operations on standard graph data structures.

in practice, because they avoid the overhead of chasing pointers and computing
hash functions; they’re just contiguous blocks of memory.

Similarly, why would anyone use linked lists in an adjacency list structure to
store neighbors, instead of balanced binary search trees or hash tables? Although
the primary reason in practice is almost surely tradition —If they were good
enough for Donald Knuth’s code, they should be good enough for yours!—there
are more principled arguments. One is that standard adjacency lists are in fact
good enough for most applications. Most standard graph algorithms never (or
rarely) actually ask whether an arbitrary edge is present or absent, or attempt
to insert or delete edges, and so optimizing the data structures to support those
operations is unnecessary.

But in my opinion, the most compelling reason for both standard data
structures is that many graphs are implicitly represented by adjacency matrices
and standard adjacency lists. For example:

• Intersection graphs are usually represented as a list of the underlying
geometric objects. As long as we can test whether two objects intersect in
constant time, we can apply any graph algorithm to an intersection graph by
pretending that the input graph is stored explicitly as an adjacency matrix.

• Any data structure composed from records with pointers between them can
be seen as a directed graph. Graph algorithms can be applied to these data
structures by pretending that the graph is stored in a standard adjacency list.

• Similarly, we can apply any graph algorithm to a configuration graph as
though it were represented as a standard adjacency list, provided we can
enumerate all possible moves from a given configuration in constant time
each.

For the last two examples, we can enumerate the edges leaving any vertex in
time proportional to its degree, but we cannot necessarily determine in constant
time if two vertices are adjacent. (Is there a pointer from this record to that
record? Can we get from this configuration to that configuration in one move?)
Moreover, we usually don’t have the luxury of reorganizing the pointers in each
record or the moves out of a given configuration into a more efficient data

196

5 - 5 - Whatever-First Search

structure. Thus, a standard adjacency list, with neighbors stored in linked lists,
is the appropriate model data structure.

In the rest of this book, unless explicitly stated otherwise, all time bounds
for graph algorithms assume that the input graph is represented by a stan¬
dard adjacency list. Similarly, unless explicitly stated otherwise, when an
exercise asks you to design and analyze a graph algorithm, you should assume
that the input graph is represented in a standard adjacency list.

5.5 Whatever-First Search

So far we have only discussed local operations on graphs; arguably the most
fundamental global question we can ask about graphs is reachability. Given a
graph G and a vertex s in G, the reachability question asks which vertices are
reachable from s; that is, for which vertices v is there a path from s to v? For
now, let’s consider only undirected graphs; I’ll consider directed graphs briefly
at the end of this section. For undirected graphs, the vertices reachable from s
are precisely the vertices in the same component as s.

Perhaps the most natural reachability algorithm—at least for people like us
who are used to thinking recursively—is depth-first search. This algorithm can
be written either recursively or iteratively. It’s exactly the same algorithm either
way; the only difference is that we can actually see the “recursion” stack in the
non-recursive version.

RecursiveDFS(v):
if v is unmarked
mark v

for each edge vw

RecursiveDFS(w)

IterativeDFS(s):

Push(s)

while the stack is not empty
v <— Pop
if v is unmarked
mark v

for each edge vw
Push(w)

Depth-first search is just one (perhaps the most common) species of a
general family of graph traversal algorithms that I call whatever-first search.
The generic traversal algorithm stores a set of candidate edges in some data
structure that I’ll call a “bag”. The only important properties of a “bag” are that
we can put stuff into it and then later take stuff back out. A stack is a particular
type of bag, but certainly not the only one. Here is the generic algorithm:

197

5. Basic Graph Algorithms

WhateverFirstSearch(s):
put s into the bag
while the bag is not empty
take v from the bag
if v is unmarked
mark v

for each edge vw

put w into the bag

I claim that WhateverFirstSearch marks every node reachable from s
and nothing else. The algorithm clearly marks each vertex in G at most once. To
show that it visits every node in a connected graph at least once, we modify the
algorithm slightly; the modifications are in bold red. Instead of keeping vertices
in the bag, the modified algorithm stores pairs of vertices. This modification
allows us to remember, whenever we visit a vertex v for the first time, which
previously-visited neighbor vertex put v into the bag. We call this earlier vertex
the parent of v.

WhateverFirstSearch(s):
put (0,s) in bag
while the bag is not empty

take (p, v) from the bag

(*)

if v is unmarked

mark v

parent(v) <— p

for each edge vw

(t)

put (v, w) into the bag

(**)

Lemma 5.1. WhateverFirstSearch(s ) marks every vertex reachable from s
and only those vertices. Moreover, the set of all pairs ( v,parent(v )) with
parent{v) ^ 0 defines a spanning tree of the component containings.

Proof: First we argue that the algorithm marks every vertex v that is reachable
from s, by induction on the shortest-path distance from s to v. The algorithm

marks s. Let v be any other vertex reachable from s, and let s->- >u->v be

any path from s to v with the minimum number of edges. (There must be such

a path, because v is reachable from s.) The prefix path s-> - m is shorter

than the shortest path from s to u, so the inductive hypothesis implies that the
algorithm marks u. When the algorithm marks u, it must immediately put the
pair (u, v) into the bag, so it must later take (tz, v) out of the bag, at which point
the algorithm immediately marks v, unless it was already marked.

Every pair ( v,parent{v )) with parent(v) 7^ 0 is actually an edge in the
underlying graph G. We claim that for any marked vertex v, the path of parent
edges v^>parent{y)^>parent{parent(y))^ ■ ■ ■ eventually leads back to s; we prove
this claim by induction on the order in which vertices are marked. Trivially s is

198

5.6. Important Variants

reachable from s, so let v be any other marked vertex. The parent of v must be
marked before v is marked, so the inductive hypothesis implies that the parent
pathparent(v)^parent(parent(v))-> • • • leads to s; adding one more parent edge
s^parent(s) establishes the claim.

The previous claim implies that every vertex marked by the algorithm is
reachable from s, and that the set of all parent edges forms a connected graph.
Because every marked node except s has a unique parent, the number of parent
edges is exactly one less than the number of marked vertices. We conclude that
the parent edges form a tree. □

Analysis

The running time of the traversal algorithm depends on what data structure we
use for the “bag”, but we can make a few general observations. Let T is the time
required to insert a single item into the bag or delete a single item from the bag.
The for loop (t) is executed exactly once for each marked vertex, and therefore
at most V times. Each edge uv in the component of s is put into the bag exactly
twice; once as the pair (u, v) and once as the pair (v, u), so line (**) is executed
at most 2 E times. Finally, we can’t take more things out of the bag than we put
in, so line (*) is executed at most 2 E + 1 times. Thus, assuming the underlying
graph G is stored in a standard adjacency list, WhateverFirstSearch runs in
0 (V + ET) time. (If G is stored in an adjacency matrix, the running time of
WhateverFirstSearch increases to 0 (V 2 + ET ).)

5.6 Important Variants

Stack: Depth-First

If we implement the “bag” using a stack, we recover our original depth-first
search algorithm. Stacks support insertions (push) and deletions (pop) in 0 ( 1 )
time each, so the algorithm runs in 0 (V + £) time. The spanning tree formed
by the parent edges is called a depth-first spanning tree. The exact shape
of the tree depends on the start vertex and on the order that neighbors are
visited inside the for loop (t), but in general, depth-first spanning trees are long
and skinny. We will consider several important properties and applications of
depth-first search in Chapter 6.

Queue: Breadth-First

If we implement the “bag” using a queue, we get a different graph-traversal
algorithm called breadth-first search. Queues support insertions (push) and
deletions (pull) in 0 ( 1 ) time each, so the algorithm runs in 0 (V + E ) time. In

199

5. Basic Graph Algorithms

this case, the breadth-first spanning tree formed by the parent edges contains
shortest paths from the start vertex s to every other vertex in its component;
we will consider shortest paths in detail in Chapter 8. Again, the exact shape of
a breadth-first spanning tree depends on the start vertex and on the order that
neighbors are visited in the for loop (t), but in general, breadth-first spanning
trees are short and bushy.

Figure 5.12. A depth-first spanning tree and a breadth-first spanning tree of the same graph, both
starting at the center vertex.

Priority Queue: Best-First

Finally, if we implement the “bag” using a priority queue, we get yet another
family of algorithms called best-first search. Because the priority queue stores
at most one copy of each edge, inserting an edge or extracting the minimum-
priority edge requires O(logE) time, which implies that best-first search runs in

0(V + £log£) time.

I describe best-first search as a “family of algorithms”, rather than a single
algorithm, because there are different methods to assign priorities to the edges,
and these choices lead to different algorithmic behavior. I’ll describe three
well-known variants below, but there are many others. In all three examples,
we assume that every edge izv or u^v in the input graph has a non-negative
weight w(uv) or w(u->v).

First, if the input graph is undirected and we use the weight of each edge
as its priority, best-first search constructs the minimum spanning tree of the
component ofs. Surprisingly, as long as all the edge weights are distinct, the
resulting tree does not depend on the start vertex or the order that neighbors
are visited; in this case, the minimum spanning tree is actually unique. This
instantiation of best-first search is commonly (but, as usual, incorrectly) known
as Prim’s algorithm; we’ll discuss this and other minimum-spanning-trees in
more detail in Chapter 7.

Define the length of a path to be the sum of the weights of its edges. We
can also compute shortest paths in weighted graphs using best-first search,
as follows. Every marked vertex v stores a distance dist(v). Initially we set
dist(s ) = 0 . For every other vertex v, when we set parent{v) <— p, we also set

200

5.6. Important Variants

dist(v ) <— dist(p) + w(p-rv), and when we insert the edge v-nv into the priority
queue, we use the priority dist(v) + w(v-»w). Assuming all edge weights are
positive, dist(v) is the length of the shortest path from s to v. This instantiation
of best-first search is commonly (but, as usual, strictly speaking, incorrectly)
known as Dijkstra’s algorithm; we’ll see this algorithm again in Chapter 8.

Finally, define the width of a path to be the minimum weight of any edge
in the path. A simple modification of “Dijkstra’s” best-first search algorithm
computes widest paths from s to every other reachable vertex; widest paths
are also called bottleneck shortest paths. Every marked vertex v stores a
value width(v). Initially we set width(s) = oo. For every other vertex v, when
we set parent{v ) <— p, we also set width(v ) <— min {width(p),w(p^v)}, and
when we insert the edge v->w into the priority queue, we use the priority
min{widtfr(v), w(v^w)}. Widest paths are useful in algorithms for computing
maximum flows, which (you guessed it) we’ll consider Chapter 10.

Disconnected Graphs

WhateverFirstSearch(s) only visits the vertices reachable from a single start
vertex s. To visit every vertex in G, we can use the following simple “wrapper”
function.

WFSAll(G):
for all vertices v
unmark v
for all vertices v

if v is unmarked

WhateverFirstSearch(v)

Wait, I hear you ask, why are you making this so complicated? Why not just 10
scan the vertex array?

MarkEveryVertexDuh( G):
for all vertices v
mark v

Well, sure, if you have an complete list of vertices, then you can do that, but
remember that not all graphs are represented so explicitly. 11 More importantly,
even if we do have an explicit vertex list, the order in which this naive algorithm
visits vertices is determined by their order in the data structure, not by the
abstract structure of the graph.

“This word is almost always a signal that you are missing something important.

u On the other hand, if we store a time-stamp at every node indicating the last time it was
“marked”, then we can “unmark every vertex” in 0(1) time by recording the start time of our
traversal, and a vertex is “marked” if its time stamp is later than the recorded start time.

201

5. Basic Graph Algorithms

In particular, unlike a naive scan through the vertices, WFSAll visits all
the vertices in one component, and then all the vertices in the next component,
and so on through each component of the input graph. This component-by-
component traversal allows us, for example, to count the components of a
disconnected graph using a single counter.

CquntComponents(G):
count <— 0
for all vertices v
unmark v
for all vertices v

if v is unmarked

count <— count +1
WhateverFirstSearch(v)
return count

With just a bit more work, we can record which component contains each vertex,
instead of merely marking it.

Count AndLabel( G):
count <— 0
for all vertices v
unmark v
for all vertices v

if v is unmarked

count *— count + 1
LabelOne(v, count )
return count

({Label one component ))
LabelOne(v, count):
while the bag is not empty
take v from the bag
if v is unmarked
mark v

comp(v) <— count

for each edge vw

put w into the bag

WFSAll marks every vertex once, puts every edge into the bag once, and
takes every edge out of the bag once, so the overall running time is 0 ( 1 / + ET),
where T is the time for a bag operation. In particular, if we run depth-first search
or breadth-first search at every vertex, the resulting algorithm still requires only
0 (1/ + E ) time.

Moreover, because WhateverFirstSearch computes a spanning tree of
one component, we can use WFSAll to computes a spanning forest of the entire
graph. In particular, best-first search with edge weights as priorities computes
the minimum-weight spanning forest in 0 ( 1 / + ElogE).

Shockingly, at least one extremely popular algorithms textbook claims that
this wrapper can only be used with depth-first search. 12 This claim is flatly
incorrect. In fact, the very first implementation of breadth-first search, written

12 To quote directly: “Unlike breadth-first search, whose predecessor subgraph forms a tree,
the predecessor subgraph produced by a depth-first search may be composed of several trees,
because the search may repeat from multiple sources.”

202

57 - Graph Reductions

around 1945 by Konrad Zuse in his proto-language Plankalkill, was developed for
the specific purpose of counting and labeling the components of an undirected
graph.

Directed Graphs

Whatever-first search is easy to adapt to directed graphs; the only difference is
that when we mark a vertex, we put all of its out-neighbors into the bag. In fact,
if we are using standard adjacency lists or adjacency matrices, we do not have
to change the code at all!

WhateverFirstSearch(s):
put s into the bag
while the bag is not empty
take v from the bag
if v is unmarked
mark v

for each edge v->w
put w into the bag

Our earlier proof implies that the algorithm marks every vertex reachable
from s, and the directed edges parent{v)^p define a rooted tree, with all edges
directed away from the root s. However, even if the graph is connected, we no
longer necessarily obtain a spanning tree of the graph, because reachability is
no longer symmetric.

On the gripping hand, WhateverFirstSearch does define a spanning tree
of the vertices reachable from s. Moreover, by varying the instantiation of the
“bag”, we can obtain a depth-first spanning tree, a breadth-first spanning tree, a
minimum-weight directed spanning tree, a shortest-path tree, or a widest-path
tree of those reachable vertices.

5.7 Graph Reductions

Flood Fill

One of the earliest modern examples of whatever-first search was proposed
by Edward Moore in the mid-1950s. A pixel map is a two-dimensional array
whose value represent colors; the individual entries in the array are called pixels,
an abbreviation of picture elements . 13 A connected region in a pixel map is a

13 Before the advent of modern raster display devices in the 1960 s, pixels were more commonly
known as stitches or tesserae, depending on whether they were made of thread or very small
rocks. The word pix became a standard abbreviation for picture(s) in the early 20 th century—not
long after sox became a common plural of sock —supplanting the earlier colloquialism piccy. See
also voxel (volume element), texel (texture element), and taxel (tactile element and/or badger).

203

5. Basic Graph Algorithms

connected subset of pixels that all have the same color, where two pixels are
considered adjacent if they are immediate horizontal or vertical neighbors. The
flood fill operation, commonly represented by a paint can in raster-graphics
editing software, changes every pixel in a connected region to a new color; the
input to the operation consists of the indices i and j of one pixel in the target
region and the new color.

■

■ ■

■Sf

ITT

B S.

.1.

"I

■■ ■

1

1

■■■

1

Figure 5.13. An example of flood fill

The flood-fill problem can be reduced to the reachability problem by chasing
the definitions. We define an undirected graph G = (V,E), whose vertices are
the individual pixels, and whose edges connect neighboring pixels with the
same color. Each connected region in the pixel map is a component of G; thus,
the flood-fill problem reduces to a reachability problem in G. We can solve this
reachability problem using whatever-first search in G, starting at the given pixel
(i, j), with one minor modification; whenever we mark a vertex, we immediately
change its color. For an n x n pixel map, the graph G has n 2 vertices and at
most 2 n 2 edges, so whatever-first search runs in OfV + E) = 0 (n 2 ) time.

This simple example demonstrates the essential ingredients of a reduction.
Rather than solving the flood-fill problem from scratch, we use an existing
algorithm as a black-box subroutine. How whatever-first search works is utterly
irrelevant here; all that matters is its specification : Given a graph G and a
starting vertex s, mark every vertex in G that is reachable from s. Like any other
subroutine, we still have to describe how to construct the input and how to
use its output. We also have to analyze our resulting algorithm in terms of our
input parameters, not the vertices and edges of whatever intermediate graph
our algorithm constructs.

Now that we have an algorithm that works—but only now—we can apply
two easy optimizations to make it faster, one practical and the other theoretical:

• In an actual implementation, we would not actually build a separate graph
data structure for G. Instead, we can use the pixel map directly as though it
were a standard adjacency list, because we can list the same-color neighbors
of any pixel in 0 ( 1 ) time each. In particular, there is no need to separately
“mark” vertices; we can use the color of the pixels instead.

• More careful analysis implies that the running time is proportional to
the number of pixels in the region being filled—equivalently, the number

204

Exercises

of vertices in component of G containing vertex (/, j)—which could be
considerably smaller than 0(n 2 ).

Exercises

Graphs

1. Prove that the following definitions are all equivalent.

• A tree is a connected acyclic graph.

• A tree is one component of a forest. (A forest is an acyclic graph.)

• A tree is a connected graph with at most V — l edges.

• A tree is a minimally connected graph; removing any edge makes the
graph disconnected.

• A tree is an acyclic graph with at least V — 1 edges.

• A tree is a maximally acyclic graph; adding an edge between any two
vertices creates a cycle.

2. Prove that any connected acyclic graph with n > 2 vertices has at least
two vertices with degree 1 . Do not use the words “tree” or “leaf”, or any
well-known properties of trees; your proof should follow entirely from the
definitions of “connected” and “acyclic”.

3. A graph ( [V,E ) is bipartite if the vertices V can be partitioned into two
subsets L and R, such that every edge has one vertex in L and the other in R.

(a) Prove that every tree is a bipartite graph.

(b) Describe and analyze an efficient algorithm that determines whether a
given undirected graph is bipartite.

4. Whenever groups of pigeons gather, they instinctively establish a pecking
order. For any pair of pigeons, one pigeon always pecks the other, driving
it away from food or potential mates. The same pair of pigeons always
chooses the same pecking order, even after years of separation, no matter
what other pigeons are around. Surprisingly, the overall pecking order can
contain cycles—for example, pigeon i pecks pigeon j, which pecks pigeon k,
which pecks pigeon l, which pecks pigeon i.

(a) Prove that any finite population of pigeons can be placed in a procession
(perhaps a parade?) so that each pigeon pecks the preceding pigeon’s
posterior. Pretty please.

205

5. Basic Graph Algorithms

(b) Suppose you are given a directed graph representing the pecking re¬
lationships among a set of n pigeons. The graph contains one vertex
per pigeon, and it contains an edge i-*j if and only if pigeon i pecks
pigeon j. Describe and analyze an algorithm to compute a pecking order
for the pigeons, as guaranteed by part (a).

(c) Prove that for any set of at least three pigeons, either the pecking order
described in part (a) is unique, or there are three pigeons i, j, and k,
such that pigeon i pecks pigeon j, which pecks pigeon k, which pecks
pigeon i.

5. An Euler tour of a graph G is a closed walk through G that traverses every
edge of G exactly once.

(a) Prove that if a connected graph G has an Euler tour, then every vertex
in G has even degree. (Euler proved this.)

(b) Prove that if every vertex in a connected graph G has even degree, then G
has an Euler tour. (Euler did not prove this.)

(c) Describe and analyze an algorithm to compute an Euler tour in a given
graph, or correctly report that no such tour exists. (Euler vaguely waved
his hands at this.)

6. The d-dimensional hypercube is the graph defined as follows. There are
2 d vertices, each labeled with a different string of d bits. Two vertices are
joined by an edge if their labels differ in exactly one bit.

(a) A Hamiltonian cycle in a graph G is a cycle of edges in G that visits every
vertex of G exactly once. Prove that for all d > 2 , the d-dimensional
hypercube has a Hamiltonian cycle.

(b) Which hypercubes have an Euler tour (a closed walk that traverses every
edge exactly once)? [Hint: This is very easy.]

Traversal Algorithms

7. Recall that a directed graph G is strongly connected if, for any two vertices u
and v, there is a path in G from u to v and a path in G from v to u.

Describe an algorithm to determine, given an undirected graph G as
input, whether it is possible to direct each edge of G so that the resulting
directed graph is strongly connected.

8. Let G be a connected graph, and let T be a depth-first spanning tree of G
rooted at some node v. Prove that if T is also a breadth-first spanning tree
of G rooted at v, then G = T.

206

Exercises

9. Professors Epprich and Goodstein propose the following optimization of the
generic whatever-first search algorithm. Instead of checking whether the
vertices we take out of the bag are marked, their algorithm checks before it
even puts the vertex into the bag, thereby ensuring that each vertex is put
into the bag at most once. Their algorithm also assigns the parent of each
vertex when that vertex is marked.

EagerWFS(s):
mark s

put s into the bag
while the bag is not empty
take v from the bag
for each edge vw

if w is unmarked
mark w
parent(w ) <— v
put w into the bag

(a) Prove that EagerWFS(s) marks every node reachable from s and nothing
else. Equivalently, prove that the parent edges v^parent(v) computed
by EagerWFS(s) define a spanning tree of the component containing s.

(b) Prove that if the bag is implemented as a queue, EagerWFS is equivalent
to breadth-first search, meaning the two algorithms mark the same
vertices in the same order and construct the same spanning tree. [Hint:
What is the definition of a queue?]

(c) Prove that EagerWFS is never equivalent to depth-first search, no matter
what data structure is used as the bag (and thus, in particular, when the
bag is a stack).

Neither EagerWFS nor RecursiveDFS specify the order that edges
vw at each vertex v are considered, and different edge orders may lead
to different spanning trees. Thus, you need to argue, for some explicit
graph G, that no spanning tree of G produced by RecursiveDFS can be
constructed by EagerWFS (using any bag data structure), or vice versa.

10. One of the earliest published descriptions of whatever-first search as a
generic class of algorithms was by Edsger Dijkstra, Leslie Lamport, Alain J.
Martin, Carel S. Scholten, and Elisabeth F. M. Steffens in 1975, as part of the
design of an automatic garbage collector. Instead of maintaining marked
and unmarked vertices, their algorithm maintains a color for each vertex,
which is either white, gray, or black. As usual, in the following algorithm,
we imagine a fixed underlying graph G.

207

5. Basic Graph Algorithms

ThreeColorSearch(s):
color all nodes white
color s gray

while at least one vertex is gray
ThreeColorStep( )

ThreeColorStepQ:
v <— any gray vertex
if v has no white neighbors
color v black

else

w <— any white neighbor of v
parentfw ) <— v
color w gray

(a) Prove that ThreeColorSearch maintains the following invariant at
all times: No black vertex is a neighbor of a white vertex. [Hint: This
should be easy.]

(b) Prove that after ThreeColorSearch(s) terminates, all vertices reach¬
able from s are black, all vertices not reachable from s are white, and
that the parent edges v^parent(v) define a rooted spanning tree of the
component containing s.

[Hint: Intuitively, black nodes are “marked” and gray nodes are “in
the bag”. Unlike our formulation of WhateverFirstSearch, however,
the three-color algorithm is not required to process all edges out of a
node at the same time.]

(c) Prove that the following variant of ThreeColorSearch, which main¬
tains the set of gray vertices in a standard stack, is equivalent to
depth-first search. [Hint: The order of the last two lines of Three-
ColorStackStep matters!]

(d) Prove that the following variant of ThreeColorSearch, which main¬
tains the set of gray vertices in a standard queue, is not equivalent
to breadth-first search. [Hint: The order of the last two lines of
ThreeColorQueueStep doesn’t matter!]

208

Exercises

ThreeColorQueueSearch(s):
color all nodes white
color s gray
push s into the queue
while at least one vertex is gray
ThreeColorQueueStep( )

ThreeColorQueueStepQ:
pull v from the queue

if v has no white neighbors
color v black

else

w <— any white neighbor of v
parent{w ) <— v
color w gray

push v into the queue
push w into the queue

r (e) Now suppose that another process is adding edges to G while Three-
ColorSearch is running. These new edges could violate the color
invariant described in part (a) and therefore destroy the correctness
of the algorithm—in particular, when ThreeColorSearch terminates,
some vertices reachable from s could be white. This would be disastrous
if we are relying on “white” to mean “unreachable and therefore safe to
delete”.

However, if the other process explicitly preserves the color invariant,
we can still use the three-color algorithm to safely identify unreachable
vertices. We model the two concurrent algorithms as follows; the
either/or choice in GarbageCollect and the choice of which vertices u
and w to Mutate are entirely out of the main algorithm’s control . 14

14 This is a dramatic oversimplification of the “mark and sweep” garbage-collection algorithms
actually used in multi-threaded languages like Lua and Go. A more thorough discussion of
multi-threaded dynamic memory management is unfortunately beyond the scope of this book,
except for the First Commandment: Thou Shalt Not Roll Thine Own Garbage Collector.

209

5. Basic Graph Algorithms

Prove that GarbageCollect eventually terminates with every vertex
reachable from s colored black and every vertex not reachable from s
colored white.

r (f) Suppose instead of recoloring black vertices gray, Mutate maintains the
color invariant by coloring some white vertices gray:

MutateQ:
u <— any vertex
w <— any vertex
if uw is not an edge
add edge uw

if u is black and w is white
color w gray
if u is white and w is black

color u gray

Prove that GarbageCollect eventually terminates with s colored black,
every vertex reachable from a black vertex colored black, and every
vertex not reachable from a black vertex colored white.

Reductions

n. A number maze is an n x n grid of positive integers. A token starts in the
upper left corner; your goal is to move the token to the lower-right corner.
On each turn, you are allowed to move the token up, down, left, or right;
the distance you may move the token is determined by the number on its
current square. For example, if the token is on a square labeled 3 , then you
may move the token three steps up, three steps down, three steps left, or
three steps right. However, you are never allowed to move the token off the
edge of the board.

Describe and analyze an efficient algorithm that either returns the
minimum number of moves required to solve a given number maze, or
correctly reports that the maze has no solution.

Figure 5.14. A 5 x 5 number maze that can be solved in eight moves.

3

5

7

4

6

5

3

1

5

3

2

8

3

1

4

4

5

7

2

3

3

1

3

2

★

(3)

"s

/

6

5

2

8

3

1

4

4

5

7

2

3

3

<5

a.

J

~t

210

12. Snakes and Ladders is a classic board game, originating in India no later
than the 16th century. The board consists of an n x n grid of squares,

Exercises

numbered consecutively from 1 to n 2 , starting in the bottom left corner and
proceeding row by row from bottom to top, with rows alternating to the
left and right. Certain pairs of squares in this grid, always in different rows,
are connected by either “snakes” (leading down) or “ladders” (leading up).
Each square can be an endpoint of at most one snake or ladder.

You start with a token in cell 1 , in the bottom left corner. In each move,
you advance your token up to k positions, for some fixed constant k. If the
token ends the move at the top end of a snake, it slides down to the bottom
of that snake. Similarly, if the token ends the move at the bottom end of a
ladder, it climbs up to the top of that ladder.

Describe and analyze an algorithm to compute the smallest number of
moves required for the token to reach the last square of the grid.

Figure 5.15. A Snakes and Ladders board. Upward straight arrows are ladders; downward wavy arrows
are snakes.

13. The infamous Mongolian puzzle-warrior Vidrach Itky Leda invented the
following puzzle in the year 1473. The puzzle consists of an n x n grid
of squares, where each square is labeled with a positive integer, and two
tokens, one red and the other blue. The tokens always lie on distinct squares
of the grid. The tokens start in the top left and bottom right corners of the
grid; the goal of the puzzle is to swap the tokens.

In a single turn, you may move either token up, right, down, or left by a
distance determined by the other token. For example, if the red token is on a
square labeled 3 , then you may move the blue token 3 steps up, 3 steps left,
3 steps right, or 3 steps down. However, you may not move either token off
the grid, and at the end of a move the two tokens cannot lie on the same
square.

Describe and analyze an efficient algorithm that either returns the
minimum number of moves required to solve a given Vidrach Itky Leda

211

5. Basic Graph Algorithms

puzzle, or correctly reports that the puzzle has no solution. For example,
given the puzzle in Figure 5.16, your algorithm would return the number 5 .

14. Suppose you are given a directed graph G = (V, E ) and two vertices s and t.
Describe and analyze an algorithm to determine if there is a walk in G from s
to t (possibly repeating vertices and/or edges) whose length is divisible
by 3 .

For example, given the graph shown below, with the indicated ver¬
tices s and t, your algorithm should return True, because the walk
s-»w->y-»x-»s-nv-»t has length 6.

15. Suppose you are given a directed graph G where some edges are red and the
remaining edges are blue. Describe an algorithm to find the shortest walk
in G from one vertex s to another vertex t in which no three consecutive
edges have the same color. That is, if the walk contains two red edges in a
row, the next edge must be blue, and if the walk contains two blue edges in
a row, the next edge must be red.

For example, given the following graph as input, your algorithm should
return the integer 7 , because is the shortest legal

walk from s to t.

16. Consider a directed graph G, where each edge is colored either red, white,
or blue. A walk in G is called a French flag walk if its sequence of edge
colors is red, white, blue, red, white, blue, and so on. More formally, a walk

v o^ v i _> '- >v k is a French flag walk if, for every integer i, the edge v ; ^v i+1

is red if i mod 3 = 0 , white if i mod 3 = 1 , and blue if i mod 3 = 2 .

Describe an algorithm to find all vertices in G that can be reached from
a given vertex v through a French flag walk.

212

Exercises

17. There are n galaxies connected by m intergalactic teleport-ways. Each
teleport-way joins two galaxies and can be traversed in both directions. Also,
each teleport-way e has an associated cost of c(e) dollars, where c(e) is a
positive integer. A teleport-way can be used multiple times, but the toll must
be paid every time it is used.

Judy wants to travel from galaxy s to galaxy t, but teleportation is not
very pleasant and she would like to minimize the number of times she needs
to teleport. However, she wants the total cost to be a multiple of five dollars,
because carrying small change is not pleasant either.

(a) Describe and analyze an algorithm to compute the smallest number of
times Judy needs to teleport to travel from galaxy s to galaxy t so that
the total cost is a multiple of five dollars.

(b) Solve part (a), but now assume that Judy has a coupon that allows her
to use exactly one teleport-way for free.

18. Three Sea Shells is a solitaire game, played on a connected undirected
graph G. Initially, three tokens are placed on distinct start vertices a, b,c.
In each turn, you must move all three tokens, by moving each token along
an edge from its current vertex to an adjacent vertex. At the end of each
turn, the three tokens must lie on three different vertices. Your goal is to
move the tokens onto three goal vertices x,y,z; it does not matter which
token ends up on which goal vertex.

x y z x y z x y z x y z x y z

Figure 5.17. The initial configuration of the Three Sea Shells puzzle and the first two turns of a solution.

Describe and analyze an algorithm to determine whether a given Three
Sea Shells puzzle is solvable. Your input consists of the graph G, the start
vertices a, b, c, and the goal vertices x,y,z. Your output is a single bit: True
or False.

19. Let G be a connected undirected graph. Suppose we start with two coins on
two arbitrarily chosen vertices of G, and we want to move the coins so that
they lie on the same vertex using as few moves as possible. At every step,
each coin must move to an adjacent vertex.

(a) Describe and analyze an algorithm to compute the minimum number of
steps to reach a configuration where both coins are on the same vertex,

213

5. Basic Graph Algorithms

or to report correctly that no such configuration is reachable. The input
to your algorithm consists of a graph G = (V, E) and two vertices u, v e V
(which may or may not be distinct).

(b) Now suppose there are three coins. Describe and analyze an algorithm
to compute the minimum number of steps to reach a configuration where
both coins are on the same vertex, or to report correctly that no such
configuration is reachable.

(c) Finally, suppose there are forty-two coins. Describe and analyze an
algorithm to determine whether it is possible to move all 42 coins to the
same vertex. Again, every coin must move at every step. For full credit,
your algorithm should run in 0 (V + E) time.

20. One of my daughter’s elementary-school math workbooks 15 contains several
puzzles of the following type:

Complete each angle maze below by tracing a path from start to finish that
has only acute angles.

Start

Describe and analyze an algorithm to solve arbitrary acute-angle mazes.

You are given a connected undirected graph G, whose vertices are points
in the plane and whose edges are line segments. Edges do not intersect,
except at their endpoints. For example, a drawing of the letter X would have
five vertices and four edges, and the first maze above has r8 vertices and 21
edges. You are also given two vertices Start and Finish.

Your algorithm should return True if G contains a walk from Start to
Finish that has only acute angles, and False otherwise. Formally, a walk
through G is valid if, for any two consecutive edges u-*v-*w in the walk,
either Zuvw = n or 0 < Zuvw < n/ 2 . Assume you have a subroutine that
can determine in 0(1) time whether the angle between two given segments
is straight, obtuse, right, or acute.

2t. Suppose you are given a set of ti horizontal and vertical line segments and
two points s and t in the plane. Describe an efficient algorithm to determine
if there is a path from s to t that does not intersect any of the given line
segments.

15 Jason Batterson and Shannon Rogers, Beast Academy Math: Practice 3A, 2012. See https:
//www.beastacademy.com/resources/printables.php for several more examples.

214

Exercises

Each horizontal line segment is specified by its left and right x-coordinates
and its unique y-coordinate; similarly, each vertical line segment is specified
by its unique x-coordinate and its top and bottom y-coordinates. Finally,
the points s and t are each specified by their x- and y-coordinates.

Figure 5.1 8 . A path between two points in a maze of horizontal and vertical line segments.

22. Every cheesy romance movie has a scene where the romantic couple, after
a long and frustrating separation, suddenly see each other across a long
distance, and then slowly approach one another with unwavering eye contact
as the music rolls in and the rain lifts and the sun shines through the clouds
and the music swells and everyone starts dancing with rainbows and kittens
and chocolate unicorns and... , l6

Suppose a romantic couple—in grand computer science tradition, named
Alice and Bob—enters their favorite park at the east and west entrances and
immediately establish eye-contact. They can’t just run directly to each other;
instead, they must stay on the path that zig-zags through the park between
the east and west entrances. To maintain the proper dramatic tension, Alice
and Bob must traverse the path so that they always lie on a direct east-west
line.

We can describe the zigzag path as two arrays X[ 0 .. n] and Y [ 0 .. n ],
containing the x- and y-coordinates of the corners of the path, in order from
the southwest endpoint to the southeast endpoint. The X array is sorted in
increasing order, and T[ 0 ] = Y[n]. The path is a sequence of straight line
segments connecting these corners.

(a) Suppose T[ 0 ] = T[n] = 0 and T[i] > 0 for every other index i; that is,
the endpoints of the path are strictly below every other point on the path.
Prove that for any path P meeting these conditions, Alice and Bob can
always meet. [Hint: Describe a graph that models all possible locations
of the couple along the path. What are the vertices of this graph? What

l6 Fun fact: Damien Chazelle, the director of Whiplash and La La Land, is the son of Princeton
computer science professor and electric guitarist Bernard Chazelle.

215

5. Basic Graph Algorithms

Figure 5.19. Alice and Bob meet. Alice walks backward in step 2 , and Bob walks backward in steps 5
and 6 .

are the edges? Use the Handshake Lemma: Every graph has an even
number of vertices with odd degree.]

(b) If the endpoints of the path are not below every other vertex, Alice and
Bob might still be able to meet, or they might not. Describe an algorithm
to decide whether Alice and Bob can meet, without either breaking
east-west eye contact or stepping off the path, given the arrays X[0.. n]
and 7 [ 0 .. n] as input.

*(c) Describe an algorithm for part (b) that runs in O(n) time.

23. The famous puzzle-maker Kaniel the Dane invented a solitaire game played
with two tokens on an n x n square grid. Some squares of the grid are
marked as obstacles, and one grid square is marked as the target. In each
turn, the player must move one of the tokens from is current position as far
as possible upward, downward, right, or left, stopping just before the token
hits (1) the edge of the board, (2) an obstacle square, or (3) the other token.
The goal is to move either of the tokens onto the target square.

Figure 5.20. An instance of Kaniel the Dane's puzzle that can be solved in six moves. Circles indicate
initial token positions; black squares are obstacles; the center square is the target.

216

For example, we can solve the puzzle shown in Figure 5.20 by moving
the red token down until it hits the obstacle, then moving the green token

Exercises

left until it hits the red token, and then moving the red token left, down,
right, and up. The red token stops at the target on the 6th move because the
green token is just above the target square.

Describe and analyze an algorithm to determine whether an instance of
this puzzle is solvable. Your input consist of the integer n, a list of obstacle
locations, the target location, and the initial locations of the tokens. The
output of your algorithm is a single boolean: True if the given puzzle is
solvable and False otherwise. [Hint: Don’t forget about the time required
to construct the graph.]

r 2 4 . Rectangle Walk is a new abstract puzzle game, available for only 99^; on
Steam, iOS, Android, Xbox One, Playstation 5, Nintendo Wii U, Atari 2600,
Palm Pilot, Commodore 64, TRS-80, Sinclair ZX-i, DEC PDP-8, PLATO,
Zuse Z3, Duramesc, Odhner Arithmometer, Analytical Engine, Jacquard
Loom, Horologium Mirabile Lundense, Leibniz Stepped Reckoner, Al-Jazari’s
Robot Band, Yan Shi’s Automaton, Antikythera Mechanism, Knotted Rope,
Ishango Bone, and Pile of Rocks.

The game is played on an n x n grid of black and white squares. The player
moves a rectangle through this grid, subject to the following conditions:

• The rectangle must be aligned with the grid; that is, the top, bottom,
left, and right coordinates must be integers.

• The rectangle must fit within the n x n grid, and it must contain at least
one grid cell.

• The rectangle must not contain a black square.

• In a single move, the player can replace the current rectangle r with any
rectangle r' that either contains r or is contained in r.

Initially, the player’s rectangle is a 1 x 1 square in the upper right corner.
The player’s goal is to reach a 1 x 1 square in the bottom left corner using
as few moves as possible.

Figure 5.21 . The first five steps of a Rectangle Walk.

Describe and analyze an algorithm to compute the length of the shortest
Rectangle Walk in a given bitmap. Your input is an array M[ 1 .. n, 1 .. n],
where M[i,j] = 1 indicates a black square and M[i,j ] = 0 indicates a
white square. Assume that a valid rectangle walk exists; in particular,

217

5. Basic Graph Algorithms

M[ 1 , 1 ] = M[n, n] = 0 . For example, given the bitmap shown above, your
algorithm should return the integer 18. [Hint: Don’t forget about the time
required to construct the graph!!]

25. Racetrack (also known as Graph Racers and Vector Rally ) is a two-player
paper-and-pencil racing game that Jeff played on the bus in 5th grade. 17 The
game is played with a track drawn on a sheet of graph paper. The players
alternately choose a sequence of grid points that represent the motion of a
car around the track, subject to certain constraints explained below.

velocity

position

(0

0 )

(1

5 )

(1

0 )

(2

5 )

(2

- 1 )

(4

4 )

(3

0 )

(7

4 )

(2

1 )

(9

5 )

(1

2 )

(10

7 )

(0

3 )

(10

10 )

(-1

4 )

(9

14 )

(0

3 )

(9

17 )

(1

2 )

(10

19 )

(2

2 )

(12

21 )

(2

1 )

(14

22 )

(2

0 )

(16

22 )

(1

- 1 )

(17

21 )

(2

- 1 )

(19

20 )

(3

0 )

(22

20 )

(3

1 )

(25

21 )

Figure 5.22. A 16 -step Racetrack run, on a 25 x 25 track. This is not the shortest run on this track.

Each car has a position and a velocity, both with integer x- and y-
coordinates. A subset of grid squares is marked as the starting area, and
another subset is marked as the finishing area. The initial position of each car
is chosen by the player somewhere in the starting area; the initial velocity of
each car is always ( 0 , 0 ). At each step, the player optionally increments or
decrements either or both coordinates of the car’s velocity; in other words,
each component of the velocity can change by at most 1 in a single step. The
car’s new position is then determined by adding the new velocity to the car’s
previous position. The new position must be inside the track; otherwise, the
car crashes and that player loses the race. The race ends when the first car
reaches a position inside the finishing area.

Suppose the racetrack is represented by an n x n array of bits, where
each 0 bit represents a grid point inside the track, each 1 bit represents a
grid point outside the track, the “starting area” is the first column, and the
“finishing area” is the last column.

17 The actual game is a bit more complicated than the version described here. See http:
//harmmade.com/vectorracer/ for an excellent online version.

218

Exercises

Describe and analyze an algorithm to find the minimum number of steps
required to move a car from the starting line to the finish line of a given
racetrack.

26. A rolling die maze is a puzzle involving a standard six-sided die (a cube with
numbers on each side) and a grid of squares. You should imagine the grid
lying on a table; the die always rests on and exactly covers one square of
the grid. In a single step, you can roll the die 90 degrees around one of its
bottom edges, moving it to an adjacent square one step north, south, east,
or west.

Some squares in the grid may be blocked; the die can never rest on a
blocked square. Other squares may be labeled with a number; whenever the
die rests on a labeled square, the number on the top face of the die must
equal the label. Squares that are neither labeled nor marked are free. You
may not roll the die off the edges of the grid. A rolling die maze is solvable
if it is possible to place a die on the lower left square and roll it to the upper
right square under these constraints.

Figure 5.23. Rolling a die

Figure 5.24 shows four rolling die mazes. Assuming we use a standard
die with 1 and 6 on opposite sides, only the first two mazes are solvable.
For example, the first maze is solvable by by placing the die on the lower
left square with 1 on the top face, and then rolling the die east, then north,
then north, then east.

Figure 5.24. Four rolling die mazes; only the first two are solvable.

■

1

1

■

(a) Suppose the input is a two-dimensional array L[ 1 .. n, 1 .. n], where each
entry L[t, j] stores the label of the square in the ith row and jth column,
where 0 means the square is free and —1 means the square is blocked.
Describe and analyze a polynomial-time algorithm to determine whether
the given rolling die maze is solvable.

r (b) Now suppose the maze is specified implicitly by a list of labeled and
blocked squares. Specifically, suppose the input consists of an integer M,

5. Basic Graph Algorithms

specifying the height and width of the maze, and an array S[ 1 .. n],
where each entry S[t] is a triple (x,y, L) indicating that square (x,y)
has label L. As in the explicit encoding, label —1 indicates that the
square is blocked; free squares are not listed in S at all. Describe and
analyze an efficient algorithm to determine whether the given rolling
die maze is solvable. For full credit, the running time of your algorithm
should be polynomial in the input size n.

[Hint: You have some freedom in how to place the initial die. There are
rolling die mazes that can be solved only if the initial position is chosen
correctly.]

*2 7. Suppose you are given an arbitrary directed graph G in which each edge is
colored either red or blue, along with two special vertices s and t.

(a) Describe an algorithm that either computes a walk from s to t such that
the pattern of red and blue edges along the walk is a palindrome, or
correctly reports that no such walk exists.

(b) Describe an algorithm that either computes the shortest walk from s
to t such that the pattern of red and blue edges along the walk is a
palindrome, or correctly reports that no such walk exists.

[Hint: Where did we last see palindromes?]

* r 28. Draughts, also known in the US as “checkers”, is a game played on an m x m
grid of squares, alternately colored light and dark. 18 The game is usually
played on an 8 x 8 or 10 x 10 board, but the rules easily generalize to any
board size. Each dark square is occupied by at most one game piece (usually
called a checker in the U.S.), which is either black or white; light squares
are always empty. One player (“White”) moves the white pieces; the other
(“Black”) moves the black pieces. A player loses when her last piece is taken
off the board.

Consider the following simple version of the game, essentially American
checkers or British draughts, but where every piece is a king. 19 Pieces can
be moved in any of the four diagonal directions. On each turn, a player
either moves one of her pieces one step diagonally into an empty square,

lS The counting tables used by medieval English government accountants were covered by a
green cloth with black squares in a checker pattern; disk-shaped counters were placed in these
squares to represent values. For this reason, the British government’s accountants have been
collectively known since the 10th century as the Exchequer. The actual counting tables were used
by the Exchequer to tally tax payments well into the 19th century.

19 Most other variants of draughts have “flying kings”, which behave very differently than kings
in the British/American game, and which make this problem much more difficult, as we will see
in Chapter 12.

220

Exercises

or makes a series of jumps with one of her pieces. In each jump, the piece
moves to an empty square two steps away in any diagonal direction, but
only if the intermediate square is occupied by a piece of the opposite color;
this enemy piece is captured and immediately removed from the board. All
jumps in the same turn must be made with the same piece.

Describe an algorithm to decide whether White can capture every black
piece, thereby winning the game, in a single turn. The input consists of
the width of the board (m), a list of positions of white pieces, and a list
of positions of black pieces. For full credit, your algorithm should run
in O(n) time, where n is the total number of pieces. [Hint: The greedy
strategy—make arbitrary jumps until you get stuck—does not always find
a winning sequence of jumps even when one exists. See problem 5. Parity,
parity, parity.]

•

•

Q

•

•

0

•

•

•

•

•

•

•

0

- U

-4 <

v-

r\

wr c

j <

m

i -

Figure 5.25. White wins in one turn.

221

And, for the hous is crinkled to and fro,

And hath so queinte weyes for to go-
For hit is shapen as the mase is wroght-
Therto have I a remedie in my thoght,

That, by a clewe of twyne, as he hath goon,

The same wey he may returne anoon,

Folwing alwey the threed, as he hath come.

— Geoffrey Chaucer, The Legend of Good Women (c. 1385)

"Com'e bello il mondo e come sono brutti i iabirinti!" dissi sollevato.

“Come sarebbe bello il mondo se ci fosse una regola per girare nei Iabirinti,"
rispose il mio maestro.

["Flow beautiful the world is, and how ugly labyrinths are," I said, relieved.

“Flow beautiful the world would be if there were a procedure for moving through
labyrinths," my master replied.]

— Umberto Eco, II nome della rosa (1980)
English translation (The Name of the Rose) by William Weaver (1983)

Depth-First Search

In the previous chapter, we considered a generic algorithm—whatever-first
search—for traversing arbitrary graphs, both undirected and directed. In this
chapter, we focus on a particular instantiation of this algorithm called depth-first
search, and primarily on the behavior of this algorithm in directed graphs.
Rather than using an explicit stack, depth-first search is normally implemented
recursively as follows:

DFS(v):

if v is unmarked
mark v

for each edge v->w
DFS(w)

We can make this algorithm slightly faster (in practice) by checking whether
a node is marked before we recursively explore it. This modification ensures

223

6 . Depth-First Search

that we call DFS(v) only once for each vertex v. We can further modify the
algorithm to compute other useful information about the vertices and edges,
by introducing two black-box subroutines, PreVisit and PostVisit, which we
leave unspecified for now.

DFS(v):
mark v
PreVisit(v)
for each edge vw
if w is unmarked
parent{w ) <— v
DFS(w)
PostVisit(v)

Recall that a node w is reachable from another node v in a directed graph G —
or more simply, v can reach w —if and only if G contains a directed path from v
to w. Let reach(v) denote the set of vertices reachable from v (including v
itself). If we unmark all vertices in G, and then call DFS(v), the set of marked
vertices is precisely reach(v).

Reachability in undirected graphs is symmetric: v can reach w if and only
if w can reach v. As a result, after unmarking all vertices of an undirected
graph G, calling DFS(v) traverses the entire component of v, and the parent
pointers define a spanning tree of that component.

The situation is more subtle with directed graphs, as shown in the figure
below. Even though the graph is “connected”, different vertices can reach
different, and potentially overlapping, portions of the graph. The parent
pointers assigned by DFS(v) define a tree rooted at v whose vertices are
precisely reach(v), but this is not necessarily a spanning tree of the graph.

Figure 6.1. Depth-first trees rooted at different vertices in the same directed graph.

As usual, we can extend our reachability algorithm to traverse the entire
input graph, even if it is disconnected, using the standard wrapper function
shown on the left in Figure 6.2. Here we add a generic black-box subroutine
Preprocess to perform any necessary preprocessing for the PreVisit and
PostVisit functions.

224

6.1. Preorder and Postorder

DFSAll(G):

DFSAll(G):

Preprocess(G)

Preprocess(G)

for all vertices v

add vertex s

unmark v

for all vertices v

for all vertices v

add edge s->v

if v is unmarked

unmark v

DFS(v)

DFS(s)

Figure 6.2. Two formulations of the standard wrapper algorithm for depth-first search

Alternatively, if we are allowed to modify the graph, we can add a new
source vertex s, with edges to every other vertex in G, and then make a single
call to DFS(s), as shown on the right of Figure 6.2. Now the resulting parent
pointers always define a spanning tree of the augmented input graph, but not of
the original input graph. Otherwise, the two wrapper functions have essentially
identical behavior; choosing one or the other is entirely a matter of convenience. 1

Again, this algorithm behaves slightly differently for undirected and directed
graphs. In undirected graphs, as we saw in the previous chapter, it is easy to
adapt DFSAll to count the components of a graph; in particular, the parent
pointers computed by DFSAll define a spanning forest of the input graph,
containing a spanning tree for each component. When the graph is directed,
however, DFSAll may discover any number of “components” between 1 and V,
even when the graph is “connected”, depending on the precise structure of the
graph and the order in which the wrapper algorithm visits the vertices.

6.1 Preorder and Postorder

Hopefully you are already familiar with preorder and postorder traversals of
rooted trees, both of which can be computed using depth-first search. Similar
traversal orders can be defined for arbitrary directed graphs—even if they are
disconnected—by passing around a counter as follows:

DFSAll(G):
clock <— 0
for all vertices v
unmark v
for all vertices v

if v is unmarked

clock <— DFS(v, dock)

DFS(v, clock):
mark v

clock <—clock + 1; v.pre <— clock
for each edge v^>w
if w is unmarked
w.parent <— v
clock <— DFS(w, clock )
clock «— clock 4-1; v.post <— clock
return clock

'The equivalence of these two wrapper functions is a specific feature of depth- first search.
In particular, wrapping breadth -first search in a for-loop to visit every vertex does not yield the
same traversal order as adding a source vertex and invoking breadth-first search at s.

225

6. Depth-First Search

Equivalently, we can use our generic depth-first-search algorithm with the
following subroutines Preprocess, PreVisit, and PostVisit.

Preprocess(G):
clock <— 0

PreVisit(v):
clock <— clock + 1
v.pre <— clock

PostVisit(v):
clock <— clock + 1
v.post <— clock

This algorithm assigns v.pre (and advance the clock) just after it pushes v
onto the recursion stack, and it assigns v.post (and advance the clock) just
before it pops v off the recursion stack. It follows that for any two vertices u
and v, the intervals [ u.pre, u.post ] and [v.pre, v.post ] are either disjoint or nested.
Moreover, [u.pre, u.post ] contains [v.pre, v.post ] if and only if DFS(v) is called
during the execution of DFS(u), or equivalently, if and only if u is an ancestor
of v in the final forest of parent pointers.

After DFSAll labels every node in the graph, the labels v.pre define a
preordering of the vertices, and the labels v.post define a postordering of the
vertices. 2 With a few trivial exceptions, every graph has several different pre-
and postorderings, depending on the order that DFS considers edges leaving
each vertex, and the order that DFSAll considers vertices.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

Figure 6.3. A depth-first forest of a directed graph, and the corresponding active intervals of its vertices,
defining the preordering abfgchdlokpeinjm and the postordering dkoplhcgfbamjnie. Forest edges are
solid; dashed edges are explained in Figure 6.4.

For the rest of this chapter, we refer to v.pre as the starting time of v (or
less formally, “when v starts”), v.post as the finishing time of v (or less formally,

2 Confusingly, both of these orders are sometimes called “depth-first ordering”. Please don’t
do that.

226

6.1. Preorder and Postorder

“when v finishes”), and the interval between the starting and finishing times as
the active internal of v (or less formally, “while v is active”).

Classifying Vertices and Edges

During the execution of DFSAll, each vertex v of the input graph has one of
three states:

• new if DFS(v) has not been called, that is, if clock < v.pre;

• active if DFS(v) has been called but has not returned, that is, if v.pre <
clock < v.post;

• finished if DFS(v) has returned, that is, if v.post < clock.

Because starting and finishing times correspond to pushes and pops on the
recursion stack, a vertex is active if and only if it is on the recursion stack. It
follows that the active nodes always comprise a directed path in G.

The edges of the input graph fall into four different classes, depending on
how their active intervals intersect. Fix your favorite edge u->v.

• If v is new when DFS(u) begins, then DFS(v) must be called during the
execution of DFS(u), either directly or through some intermediate recursive
calls. In either case, u is a proper ancestor of v in the depth-first forest, and
u.pre < v.pre < v.post < u.post.

- If DFS(iz) calls DFS(v) directly, then u = v.parent and u^v is called a
tree edge.

- Otherwise, u->v is called a forward edge.

• If v is active when DFS(iz) begins, then v is already on the recursion stack,
which implies the opposite nesting order v.pre < u.pre < u.post < v.post.
Moreover, G must contain a directed path from v to u. Edges satisfying this
condition are called back edges.

• If v is finished when DFS(u) begins, we immediately have v.post < u.pre.
Edges satisfying this condition are called cross edges.

• Finally, the fourth ordering u.post < v.pre is impossible.

These edge classes are illustrated in Figure 6.4. Again, the actual classification
of edges depends on the order in which DFSAll considers vertices and the order
in which DFS considers the edges leaving each vertex.

Finally, the following key lemma characterizes ancestors and descendants in
any depth-first forest according to vertex states during the traversal.

Lemma 6.1. Fix an arbitrary depth-first traversal of any directed graph G. The
following statements are equivalent for all vertices u and v of G.

(a) u is an ancestor ofv in the depth-first forest.

227

6 . Depth-First Search

tree ! forward “. . cro f5--II I

u ; back dr¬

ill wm □

Figure 6.4. Classification of edges by depth-first search.

(b) u.pre < v.pre < v.post < u.post.

(c) Just after DFS(v) is called, u is active.

(d) Just before DFS(u) is called, there is a path from u to v in which every
vertex (including u and v) is new.

Proof: First, suppose u is an ancestor of v in the depth-first forest. Then by
definition there is a path P of tree edges u to v. By induction on the path
length, we have u.pre < w.pre < w.post < u.post for every vertex w in P, and
thus every vertex in P is new before DFS(u) is called. In particular, we have

u. pre < v.pre < v.post < u.post, which implies that u is active while DFS(v) is
executing.

Because parent pointers correspond to recursive calls, u.pre < v.pre <

v. post < u.post implies that u is an ancestor of v.

Suppose u is active just after DFS(v) is called. Then u.pre < v.pre < v.post <
u.post, which implies that there is a path of (zero or more) tree edges from u,
through the intermediate nodes on the recursion stack (if any), to v.

Finally, suppose u is not an ancestor of v. Fix an arbitrary path P from u
to v, let x be the first vertex in P that is not a descendant of u, and let w be
the predecessor of x in P. The edge w->x guarantees that x.pre < w.post, and

w. post < u.post because w is a descendant of u, so x.pre < u.post. It follows that

x. pre < u.pre, because otherwise x would be a descendant of u. Because active
intervals are properly nested, there are only two possibilities:

• If u.post < x.post, then x is active when DFS(u) is called.

• If x.post < u.pre, then x is already finished when DFS(iz) is called.

We conclude that every path from u to v contains a vertex that is not new when
DFS(u) is called. □

228

6.2. Detecting Cycles

6.2 Detecting Cycles

A directed acyclic graph or dag is a directed graph with no directed cycles.
Any vertex in a dag that has no incoming vertices is called a source; any vertex
with no outgoing edges is called a sink. An isolated vertex with no incident
edges at all is both a source and a sink. Every dag has at least one source and
one sink, but may have more than one of each. For example, in the graph with
n vertices but no edges, every vertex is a source and every vertex is a sink.

Figure 6.5. A directed acyclic graph. Vertices e, f, and j are sources; vertices b, c, and p are sinks.

Recall from our earlier case analysis that if u.post < v.post for any edge u^v,
the graph contains a directed path from v to u, and therefore contains a directed
cycle through the edge u->v. Thus, we can determine whether a given directed
graph G is a dag in 0 (V + E ) time by computing a postordering of the vertices
and then checking each edge by brute force.

Alternatively, instead of numbering the vertices, we can explicitly maintain
the status of each vertex and immediately return False if we ever discover
an edge to an active vertex. This algorithm also runs in 0 {V + E) time; see
Figure 6.6.

IsAcyclic(G):
for all vertices v

v.status <— New
for all vertices v

if v.status = New

if IsAcyclicDFS(v) = False
return False

return True

IsAcYCLicDFSfv):
v.status <— Active
for each edge v^w

if w.status = Active
return False
else if w.status = New

if IsAcyclicDFS(w) = False
return False
v.status <— Finished
return True

Figure 6.6. A linear-time algorithm to determine if a graph is acyclic.

229

6 . Depth-First Search

6.3 Topological Sort

A topological ordering of a directed graph G is a total order -< on the vertices
such that u < v for every edge u->v. Less formally, a topological ordering
arranges the vertices along a horizontal line so that all edges point from left to
right. A topological ordering is clearly impossible if the graph G has a directed
cycle—the rightmost vertex of the cycle would have an edge pointing to the left!

On the other hand, consider an arbitrary postordering of an arbitrary
directed graph G. Our earlier analysis implies that u.post < v.post for any edge
u->v, then G contains a directed path from v to u, and therefore contains a
directed cycle through u->v. Equivalently, if G is acyclic, then u.post > v.post for
every edge u->v. It follows that every directed acyclic graph G has a topological
ordering; in particular, the reversal of any postordering of G is a topological
ordering of G.

If we require the topological ordering in a separate data structure, we can
simply write the vertices into an array in reverse postorder, in 0 (V + E) time, as
shown in Figure 6.8.

Implicit Topological Sort

But recording the topological order into a separate data structure is usually
overkill. In most applications of topological sort, the ordered list of the vertices
is not our actual goal; rather, we want to perform some fixed computation at
each vertex of the graph, either in topological order or in reverse topological
order. For these applications, it is not necessary to record the topological order
at all!

230

6 .3- Topological Sort

TopologicalSort(G):
for all vertices v
v.status <— New
clock <— V
for all vertices v

if v.status = New

clock <— TopSortDFS(v, clock )
return S[1.. V]

TqpSortDFS(v, clock):
v.status <— Active
for each edge v^w
if w.status = New

clock <— TopSortDFS(v, clock )
else if w.status = Active
fail gracefully
v.status <— Finished
S [clock] «— v
clock <— clock — 1
return clock

Figure 6.8. Explicit topological sort

If we want to process a directed acyclic graph in reverse topological order,
it suffices to process each vertex at the end of its recursive depth-first search.
After all, topological order is the same as reversed postorder!

PqstPrqcess(G):
for all vertices v

v.status <— New
for all vertices v

if v is unmarked

PostProcessDFS(v)

PostPrqcessDFS(v):
v.status <— Active
for each edge v^w
if w.status = New

PostProcessDFS(w)
else if w.status = Active
fail gracefully
v.status <— Finished
Process(v)

If we already know that the input graph is acyclic, we can further simplify the
algorithm by simply marking vertices instead of recording their search status.

PostPrqcessDag(G):
for all vertices v
unmark v
for all vertices v

if v is unmarked

PostProcessDagDFS(s)

PostProcessDagDFS(v):
mark v

for each edge v->w
if w is unmarked

PostProcessDagDFS(w)

Process(v)

This is just the standard depth-first search algorithm, with PostVisit renamed
to Process!

Because it is such a common operation on directed acyclic graphs, I sometimes
express postorder processing of a dag idiomatically as follows:

PqstPrqcessDag(G):
for all vertices v in postorder
Process(v)

231

6 . Depth-First Search

For example, our earlier explicit topological sort algorithm can be written as
follows:

TopologicalSort(G):
clock <— V

for all vertices v in postorder
S [clock] <— v
clock <— clock — 1
return S[1 ..V]

To process a dag in forward topological order, we can record a topological
ordering of the vertices into an array and then run a simple for-loop. Alternatively,
we can apply depth-first search to the reversal of G, denoted rev(G), obtained
by replacing each each v-nv with its reversal w-tv. Reversing a directed cycle
gives us another directed cycle with the opposite orientation, so the reversal
of a dag is another dag. Every source in G is a sink in rev(G) and vice versa; it
follows inductively that every topological ordering of rev(G) is the reversal of a
topological ordering of G. 3 The reversal of any directed graph (represented in a
standard adjacency list) can be computed in 0 (V + E) time; the details of this
construction are left as an easy exercise.

6.4 Memoization and Dynamic Programming

Our topological sort algorithm is arguably the model for a wide class of dynamic
programming algorithms. Recall that the dependency graph of a recurrence
has a vertex for every recursive subproblem and an edge from one subproblem
to another if evaluating the first subproblem requires a recursive evaluation
of the second. The dependency graph must be acyclic, or the naive recursive
algorithm would never halt.

Evaluating any recurrence with memoization is exactly the same as perform¬
ing a depth-first search of the dependency graph. In particular, a vertex of the
dependency graph is “marked” if the value of the corresponding subproblem has
already been computed. The black-box subroutines PreVisit and PostVisit
are proxies for the actual value computation. See Figure 6.9.

Carrying this analogy further, evaluating a recurrence using dynamic pro¬
gramming is the same as evaluating all subproblems in the dependency graph of
the recurrence in reverse topological order—every subproblem is considered
after the subproblems it depends on. Thus, every dynamic programming al¬
gorithm is equivalent to a postorder traversal of the dependency graph of its
underlying recurrence!

3 A postordering of the reversal of G is not necessarily the reversal of a postordering of G,
even though both are topological orderings of G.

232

6.4- Memoization and Dynamic Programming

Memoize(x) :

DFS(v) :

if value[x] is undefined

if v is unmarked

initialize vaZue[x]

mark v
PreVisit(x)

for all subproblems y of x

for all edges v^w

MEMOIZE(y)

update vaZue[x] based on value[y]

DFS(w)

finalize value[x ]

PostVisit(x)

Figure 6.9. Memoized recursion is depth-first search. Depth-first search is memoized recursion.

DynamicProgramming(G) :
for all subproblems a in postorder
initialize value[x]
for all subproblems y of x

update value[x] based on value[y]
finalize value[x]

Figure 6.10. Dynamic programming is postorder traversal.

However, there are some minor differences between most dynamic program¬
ming algorithms and topological sort. First, in most dynamic programming
algorithms, the dependency graph is implicit —the nodes and edges are not
explicitly stored in memory, but rather are encoded by the underlying recur¬
rence. But this difference really is minor; as long as we can enumerate recursive
subproblems in constant time each, we can traverse the dependency graph
exactly as if it were explicitly stored in an adjacency list.

More significantly, most dynamic programming recurrences have highly
structured dependency graphs. For example, as we discussed in Chapter 5,
the dependency graph for the edit distance recurrence is a regular grid with
diagonals, and the dependency graph for optimal binary search trees is an
upper triangular grid with all possible rightward and upward edges. This
regular structure allows us to hard-wire a (reverse) topological order directly
into the algorithm, typically as a collection of nested loops, so there is no need
to topologically sort the dependency graph at run time. We previously called
the reverse topological order an evaluation order.

Dynamic Programming in Dags

Conversely, we can use depth-first search to build dynamic programming
algorithms for problems with less structured dependency graphs. For example,
consider the longest path problem, which asks for the path of maximum total
weight from one node s to another node t in a directed graph G with weighted
edges. The longest path problem is NP-hard in general directed graphs, by an

233

6. Depth-First Search

Figure 6.11. The dependency dag of the edit distance recurrence.

easy reduction from the traveling salesman problem, but it is easy to solve in
linear time if the input graph G is acyclic, as follows.

Fix the target vertex t, and for any node v, let LLP(v) denote the Length
of the Longest Path in G from v to t. If G is a dag, this function satisfies the
recurrence

LLP(v)

1 °

|ma xjf(v->w)+ LLP(w) v^wGF}

if v = t,
otherwise,

where l (v^w) denotes the given weight (“length”) of edge v-nv, and max0 =
—00. In particular, if v is a sink but not equal to t, then LLP(v) = —00.

The dependency graph for this recurrence is the input graph G itself:
subproblem LLP(v) depends on subproblem LLP(w) if and only if v-nv is an
edge in G. Thus, we can evaluate this recursive function in 0 (V + E) time by
performing a depth-first search of G, starting at s. The algorithm memoizes
each length LLP(v) into an extra field in the corresponding node v.

LongestPath(v, t):
if v = t

return 0

if v.LLP is undefined

v.LLP <-00

for each edge v->w

v.LLP <— max {v.LLP, £(v^w) + LongestPath(w, t)}
return v.LLP

In principle, we can transform this memoized recursive algorithm into a
dynamic programming algorithm via topological sort:

234

6 . 5 . Strong Connectivity

LongestPath(s, t):
for each node v in postorder
if v = t

v.LLP <- 0

else

v.LLP * -oo

for each edge v->w

v.LLP <— max {v.LLP, £(v^w) + w.LLP}

return s.LLP

These two algorithms are arguably identical—the recursion in the first algorithm
and the for-loop in the second algorithm represent the “same” depth-first
search! Choosing one of these formulations over the other is entirely a matter
of convenience.

Almost any dynamic programming problem that asks for an optimal sequence
of decisions can be recast as finding an optimal path in some associated dag. For
example, the text segmentation, subset sum, longest increasing subsequence,
and edit distance problems we considered in Chapters 2 and 3 can all be
reformulated as finding either a longest path or a shortest path in a dag, possibly
with weighted vertices or edges. On the other hand, “tree-shaped” dynamic
programming problems, like finding optimal binary search trees or maximum
independent sets in trees, cannot be recast as finding an optimal path in a dag.

6.5 Strong Connectivity

Let’s go back to the proper definition of connectivity in directed graphs. Recall
that one vertex u can reach another vertex v in a directed graph G if G contains
a directed path from u to v, and that reach{u ) denotes the set of all vertices
that u can reach. Two vertices u and v are strongly connected if u can reach v
and v can reach u. A directed graph is strongly connected if and only if every
pair of vertices is strongly connected.

Tedious definition-chasing implies that strong connectivity is an equivalence
relation over the set of vertices of any directed graph, just like connectivity in
undirected graphs. The equivalence classes of this relation are called the strongly
connected components —or more simply, the strong components —of G. Equiv¬
alently, a strong component of G is a maximal strongly connected subgraph
of G. A directed graph G is strongly connected if and only if G has exactly one
strong component; at the other extreme, G is a dag if and only if every strong
component of G consists of a single vertex.

The strong component graph scc(G) is another directed graph obtained
from G by contracting each strong component to a single vertex and collapsing
parallel edges. (The strong component graph is sometimes also called the

235

6. Depth-First Search

meta-graph or condensation of G.) It’s not hard to prove (hint, hint) that scc(G)
is always a dag. Thus, at least in principle, it is possible to topologically order
the strong components of G; that is, the vertices can be ordered so that every
back edge joins two edges in the same strong component.

Figure 6.12. The strong components of a graph G and the strong component graph scc(G).

It is straightforward to compute the strong component of a single vertex v
in 0 ( 1 / + E) time. First we compute reach(v) via whatever-first search. Then
we compute reach -1 (v) = {u \ v e reach(u)} by searching the reversal of G.
Finally, the strong component of v is the intersection reach(v) n reach -1 (v). In
particular, we can determine whether the entire graph is strongly connected in
0 (1/ + E ) time.

Similarly, we can compute all the strong components in a directed graph
by combining the previous algorithm with our standard wrapper function.
However, the resulting algorithm runs in 0 (VE ) time; there are at most V strong
components, and each requires 0 (E) time to discover, even when the graph is a
dag. Surely we can do better! After all, we only need 0 ( 1 / + E) time to decide
whether every strong component is a single vertex.

6.6 Strong Components in Linear Time

In fact, there are several algorithms to compute strong components in 0 ( 1 / + E)
time, all of which rely on the following observation.

Lemma 6.2. Fix a depth-first traversal of any directed graph G. Each strong
component C of G contains exactly one node that does not have a parent in C.
(Either this node has a parent in another strong component, or it has no parent.)

Proof: Let C be an arbitrary strong component of G. Consider any path from
one vertex v e C to another vertex weC. Every vertex on this path can reach
w, and thus can reach every vertex in C; symmetrically, every node on this path
can be reached by v, and thus can be reached by every vertex in C. We conclude
that every vertex on this path is also in C.

236

6 .6. Strong Components in Linear Time

Let v be the vertex in C with the earliest starting time. If v has a parent,
then parent(y) starts before v and thus cannot be in C.

Now let w be another vertex in C. Just before DFS(v) is called, every vertex
in C is new, so there is a path of new vertices from v to w. Lemma 1 now implies
that w is a descendant of v in the depth-first forest. Every vertex on the path of
tree edges v to w lies in C; in particular, parent(w ) € C. □

The previous lemma implies that each strong component of a directed
graph G defines a connected subtree of any depth-first forest of G. In particular,
for any strong component C, the vertex in C with the earliest starting time is the
lowest common ancestor of all vertices in C; we call this vertex the root of C.

Figure 6.13. Strong components are contiguous in the depth-first forest.

I’ll present two algorithms, both of which follow the same intuitive outline.
Let C be any strong component of G that is a sink in scc(G); we call C a sink
component. Equivalently, C is a sink component if the reach of any vertex
in C is precisely C. We can find all the strong components in G by repeatedly
finding a vertex v in some sink component (somehow), finding the vertices
reachable from v, and removing that sink component from the input graph,
until no vertices remain. This isn’t quote an algorithm yet, because it’s not clear
how to find a vertex in a sink component!

StrongComponents(G):
count <— 0

while G is non-empty
C^0

count <— count + 1

v <— any vertex in a sink component of G ((Magic!))
for all vertices w in reach(v)
w.label <— count
add w to C

remove C and its incoming edges from G

Figure 6.14. Almost an algorithm to compute strong components.

237

6. Depth-First Search

Koraraju and Sharir’s Algorithm

At first glance, finding a vertex in a sink component quickly seems quite difficult.
However, it’s actually quite easy to find a vertex in a source component—a strong
component of G that corresponds to a source in scc(G)—using depth-first search.

Lemma 6.3. The last vertex in any postordering ofG lies in a source component
of G.

Proof: Fix a depth-first traversal of G, and let v be the last vertex in the resulting
postordering. Then DFS(v) must be the last direct call to DFS made by the
wrapper algorithm DFSAll. Moreover, v is the root of one of the trees in
the depth-first forest, so any node x with x.post > v.pre is a descendant of v.
Finally, v is the root of its strong component C.

For the sake of argument, suppose there is an edge x->y such that x <£. C
and y e C. Then x can reach y, and y can reach v, so x can reach v. Because v
is the root of C, vertex y is a descendant of v, and thus v.pre < y.pre. The edge
x-*y guarantees that y.pre < x.post and therefore v.pre < x.post. It follows
that x is a descendant of v. But then v can reach x (through tree edges),
contradicting our assumption that x ^ C. □

It is easy to check (hint, hint) that rev(scc(G )) = scc(rev(G)) for any directed
graph G. Thus, the last vertex in a postordering of rev(G) lies in a sink component
of the original graph G. Thus, if we traverse the graph a second time, where the
wrapper function follows a reverse postordering of rev(G), then each call to DFS
visits exactly one strong component of G. 4

Putting everything together, we obtain the algorithm shown in Figure 6.15,
which counts and labels the strong components of any directed graph in 0 (V +E)
time. This algorithm was discovered (but never published) by Rao Kosaraju
in 1978, and later independently rediscovered by Micha Sharir in 1981. 5 The
Kosaraju-Sharir algorithm has two phases. The first phase performs a depth-first
search of rev(G), pushing each vertex onto a stack when it is finished. In the
second phase, we perform a whatever- first traversal of the original graph G,
considering vertices in the order they appear on the stack. The algorithm labels
each vertex with the root of its strong component (with respect to the second
depth-first traversal).

Figure 6.16 shows the Koraraju-Sharir algorithm running on our example
graph. With only minor modifications to the algorithm, we can also compute
the strong component graph scc(G) in 0 (V + E) time.

4 Again: A reverse postordering of rev(G) is not the same as a postordering of G.

5 There are rumors that the same algorithm appears in the Russian literature even before
Kosaraju, but I haven’t found a reliable source for that rumor yet.

238

6 .6. Strong Components in Linear Time

Figure 6.15. The Kosaraju-Sharir strong components algorithm

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239

6. Depth-First Search

*Tarjan’s Algorithm

An earlier but considerably more subtle linear-time algorithm to compute strong
components was published by Bob Tarjan in 1972. 6 Intuitively, Tarjan’s algorithm
identifies a source component of G, “deletes” it, and then “recursively” finds the
remaining strong components; however, the entire computation happens during
a single depth-first search.

Fix an arbitrary depth-first traversal of some directed graph G. For each
vertex v, let low(v) denote the smallest starting time among all vertices reachable
from v by a path of tree edges followed by at most one non-tree edge. Trivially,
low(v) < v.pre, because v can reach itself through zero tree edges followed by
zero non-tree edges. Tarjan observed that sink components can be characterized
in terms of this low function.

Lemma 6 . 4 . A vertex v is the root of a sink component of G if and only if
low(v) = v.pre and Zow(w) < w.pre for every proper descendant w ofv.

Proof: First, let v be a vertex such that Zow(v) = v.pre. Then there is no edge
w->x where w is a descendant of v and x.pre < v.pre. On the other hand, v
cannot reach any vertex y such that y.pre > v.post. It follows that v can
reach only its descendants, and therefore any descendant of v can reach only
descendants of v. In particular, v cannot reach its parent (if it has one), so v is
the root of its strong component.

Now suppose in addition that Zow(w) < w.pre for every descendant w of v.
Then each descendant w can reach another vertex x (which must be another
descendant of v) such that x.pre < w.pre. Thus, by induction, every descendant
of v can reach v. It follows that the descendants of v comprise the strong
component C whose root is v. Moreover, C must be a sink component, because v
cannot reach any vertex outside of C.

On the other hand, suppose v is the root of a sink component C. Then v
can reach another vertex w if and only if w € C. But v can reach all of its
descendants, and every vertex in C is a descendant of v, so v’s descendants
comprise C. If Zow(w) = w.pre for any other node w € C, then w would be
another root of C, which is impossible. □

Computing Zow(v) for every vertex v via depth-first search is straightforward;
see Figure 6 . 17 .

Lemma 6.4 implies that after running FindLow, we can identify the root
of every sink component in 0(V + E) time (by a global whatever-first search),

According to legend, Kosaraju apparently discovered his algorithm during an algorithms
lecture; he was supposed to present Tarjan’s algorithm, but he forgot his notes, so he had to
make up something else on the fly. The only thing I find surprising about this story is that nobody
tells it about Sharir or Tarjan.

240

6 .6. Strong Components in Linear Time

FindLqw(G):
clock <— 0
for all vertices v
unmark v
for all vertices v

if v is unmarked
FindLowDFS(v)

FindLqwDFS(v):
mark v

clock <— clock + 1
v.pre <— clock
v.low <— v.pre
for each edge v^w
if w is unmarked
FindLowDFS(w)
v.Zow <— min{v. low, w.Zow}

else

v.Zow <— min{ v.low, w.pre }

Figure 6.17. Computing low[v ) for every vertex v.

and then mark and delete those sink components in 0(V + E ) additional time
[by calling whatever-first search at each root), and then recurse. Unfortunately,
the resulting algorithm might require V iterations, each removing only a single
vertex, naively giving us a total running time of 0(VE ).

To speed up this strategy, Tarjan’s algorithm maintains a stack of vertices
(separate from the recursion stack). Whenever we start a new vertex v, we push
it onto the stack. Whenever we finish a vertex v, we compare v.low with v.pre.
Then the first time we discover that v.low = v.pre, we know three things:

• Vertex v is the root of a sink component C.

• All vertices in C appear consecutively at the top of the stack.

• The deepest vertex in C on the stack is v.

At this point, we can identify the vertices in C by popping them off the stack
one by one, stopping when we pop v.

We could delete the vertices in C and recursively compute the strong
components of the remaining graph, but that would be wasteful, because we
would repeat verbatim all computation done before visiting v. Instead, we
label each vertex in C, identifying v as the root of its strong component, and
then ignore labeled vertices for the rest of the depth-first search. Formally, this
modification changes the definition of Zow(v) to the smallest starting time among
all vertices in the same strong component as v that v can reach by a path of
tree edges followed by at most one non-tree edge. But to prove correctness, it’s
easier to observe that ignoring labeled vertices leads the algorithm to exactly
the same behavior as actually deleting them.

Finally, Tarjan’s algorithm is shown in Figure 6 . 18 , with the modifications
from FindLow (Figure 6 . 17 ) indicated in bold red. The running time of the
algorithm can be split into two parts. Each vertex is pushed onto S once and
popped off S once, so the total time spent maintaining the stack (the red stuff)
is O(V). If we ignore the stack maintenance, the rest of the algorithm is just a

241

6. Depth-First Search

standard depth-first search. We conclude that the algorithm runs in 0(V + E)
time.

Tarjan(G):
clock <— 0

S <— new empty stack

for all vertices v
unmark v
v.root <— None
for all vertices v

if v is unmarked
TarjanDFS(v)

TarjanDFS(v):
mark v

clock <— clock + 1
v.pre <— dock
v.Zow <— v.pre
Push(S, v)
for each edge v->w
if w is unmarked
TarjanDFS(w)
v.Zow <— min{ v.Zow, w.Zow}
else if w.root = None

v. Zow <— min{ v.Zow, w.pre}
if v.Zow = v.pre

repeat

w <— Pop(S)

w. root <— v
until w = v

Figure 6.18. Tarjan's strong components algorithm.

Exercises

Depth-first search, topological sort, and strong components

o. (a) Describe an algorithm to compute the reversal rev(G) of a directed graph
in 0(V + E ) time.

(b) Prove that for every directed graph G, the strong component graph
scc(G) is acyclic.

(c) Prove that scc(rev(G )) = rev(scc(G)) for every directed graph G.

(d) Fix an arbitrary directed graph G. For any vertex v of G, let S(v) denote
the strong component of G that contains v. For all vertices u and v of G,
prove that v is reachable from u in G if and only if S(v) is reachable
from S(iz) in scc(G).

(e) Suppose S and T are two strong components in a directed graph G.
Prove that either finish(u) <finish(y ) for all vertices u € S and v € T, or
finish(u ) > finish{y ) for all vertices u € S and v£T.

r. A directed graph G is semi-connected if, for every pair of vertices iz and v,

either u is reachable from v or v is reachable from u (or both).

(a) Give an example of a dag with a unique source that is not semi-connected.

242

Exercises

(b) Describe and analyze an algorithm to determine whether a given directed
acyclic graph is semi-connected.

(c) Describe and analyze an algorithm to determine whether an arbitrary
directed graph is semi-connected.

2 . The police department in the city of Shampoo-Banana has made every street
in the city one-way. Despite widespread complaints from confused motorists,
the mayor claims that it is possible to legally drive from any intersection in
Shampoo-Banana to any other intersection.

(a) The city needs to either verify or refute the mayor’s claim. Formalize this
problem in terms of graphs, and then describe and analyze an algorithm
to solve it.

(b) After running your algorithm from part (a), the mayor reluctantly admits
that she was lying misinformed. Call an intersection x good if, for any
intersection y that one can legally reach from x, it is possible to legally
drive from y back to x. Now the mayor claims that over 95 % of the
intersections in Shampoo-Banana are good. Describe and analyze an
efficient algorithm to verify or refute her claim.

For full credit, both algorithms should run in linear time.

3 . A vertex v in a connected undirected graph G is called a cut vertex if the
subgraph G — v (obtained by removing v from G) is disconnected.

Figure 6.19. A dag with three cut vertices.

(a) Describe a linear-time algorithm that determines, given a graph G and
a vertex v, whether v is a cut vertex in G. What is the running time to
find all cut vertices by trying your algorithm for each vertex?

(b) Describe and analyze an algorithm that correctly determines whether
a given directed acyclic graph contains a cut vertex. [Hint: This is a
warmup.]

(c) Let T be an arbitrary spanning tree of G, rooted at an arbitrary vertex r.
For each vertex v, let T v denote the subtree of T rooted at v; for example,
T r = T. Prove that v is a cut vertex of G if and only if G does not have
an edge with exactly one endpoint in T v .

243

6. Depth-First Search

(d) Describe a linear-time algorithm to identify every cut vertex in G. [Hint:
Let T be a depth-first spanning tree ofG.]

4 . An edge e in a connected undirected graph G is called a cut edge if the
subgraph G — e (obtained by removing e from G) is disconnected.

(a) Given G and edge e describe a linear-time algorithm that determines
whether e is a cut edge or not. What is the running time to find all cut
edges by trying your algorithm for each edge?

(b) Let T be an arbitrary spanning tree of G. Prove that every cut edge of
G is also an edge in T. This claim implies that G has at most V — 1 cut
edges. How does this information improve your algorithm from part (a)
to find all cut-edges?

(c) Now suppose we root T at an arbitrary vertex r. For any vertex v, let
T v denote the subtree of T rooted at v; for example, T r = T. Let uv be
an arbitrary edge of T, where u is the parent of v. Prove that uv is a
cut edge of G if and only if uv is the only edge in G with exactly one
endpoint in T v .

(d) Describe a linear-time algorithm to identify every cut edge in G. [Hint:
Let T be a depth-first spanning tree ofG.]

5 . The transitive closure G T of a directed graph G is a directed graph with
the same vertices as G, that contains any edge u^v if and only if there is a
directed path from u to v in G. A transitive reduction of G is a graph with
the smallest possible number of edges whose transitive closure is G r . The
same graph may have several transitive reductions.

(a) Describe an efficient algorithm to compute the transitive closure of a
given directed graph.

(b) Prove that a directed graph G has a unique transitive reduction if and
only if G is acyclic.

(c) Describe an efficient algorithm to compute a transitive reduction of a
given directed graph.

6 . One of the oldest algorithms for exploring arbitrary connected graphs was
proposed by Gaston Tarry in 1895 , as a procedure for solving mazes . 7 The
input to Tarry’s algorithm is an undirected graph G; however, for ease of
presentation, we formally split each undirected edge uv into two directed
edges u—»v and v->u. (In an actual implementation, this split is trivial;

7 Even older graph-traversal algorithms were described by Charles Tremaux in 1882, by
Christian Wiener in 1873, and (implicitly) by Leonhard Euler in 1736. Wiener’s algorithm is
equivalent to depth-first search in a connected undirected graph.

244

Exercises

the algorithm simply uses the given adjacency list for G as though G were
directed.)

Tarry(G):

unmark all vertices of G
color all edges of G white
s <— any vertex in G
RecTarry(s)

RecTarry(v):

mark v

(("visit v"))

if there is a white arc

v—»w

if w is unmarked

color w-'V green

color v^w red

| (("traverse v^w"))

RecTarry(w)

else if there is a green arc v->w

color v^w red

} (("traverse v^w"))

RecTarry(w)

We informally say that Tarry’s algorithm “visits” vertex v every time

it marks v, and it “traverses” edge v->w when it colors that edge red and

recursively calls RecTarry(w). Unlike our earlier graph traversal algorithm,

Tarry’s algorithm can mark same vertex multiple times.

(a) Describe how to implement Tarry’s algorithm so that it runs in 0(V +E)
time.

(b) Prove that no directed edge in G is traversed more than once.

(c) When the algorithm visits a vertex v for the /cth time, exactly how many
edges into v are red, and exactly how many edges out of v are red?
[Hint: Consider the starting vertex s separately from the other vertices.]

(d) Prove each vertex v is visited at most deg(v) times, except the starting
vertex s, which is visited at most deg(s)+1 times. This claim immediately
implies that Tarry(G) terminates.

(e) Prove that the last vertex visited by Tarry(G) is the starting vertex s.

(f) For every vertex v that Tarry(G) visits, prove that all edges into v and
out of v are red when Tarry(G) halts. [Hint: Consider the vertices in
the order that they are marked for the first time, starting with s, and
prove the claim by induction.]

(g) Prove that Tarry(G) visits every vertex of G. This claim and the previous
claim imply that Tarry(G) traverses every edge of G exactly once.

7 . Consider the following variant of Tarry’s graph-traversal algorithm; this
variant traverses green edges without recoloring them red and assigns two
numerical labels to every vertex:

245

6. Depth-First Search

Tarry2(G):

unmark all vertices of G
color all edges of G white
s <— any vertex in G
RecTarry(s, 1)

RecTarry2(v, clock) :
if v is unmarked

v.pre <— clock; clock <— clock + 1
mark v

if there is a white arc v->w
if w is unmarked

color w->v green
color v^w red
RecTarry2(w, clock )
else if there is a green arc v->w

v.post <— clock; clock <— clock + 1
RecTarry2(w, clock )

Prove or disprove the following claim: When Tarry2(G) halts, the green
edges define a spanning tree and the labels v.pre and v.post define a preorder
and postorder labeling that are all consistent with a single depth-first search
of G. In other words, prove or disprove that Tarry 2 produces the same
output as depth-first search, even though it visits the edges in a completely
different order.

8. You have a collection of n lock-boxes and m gold keys. Each key unlocks at
most one box. However, each box might be unlocked by one key, by multiple
keys, or by no keys at all. There are only two ways to open each box once it
is locked: Unlock it properly (which requires having one matching key in
your hand), or smash it to bits with a hammer.

Your baby brother, who loves playing with shiny objects, has somehow
managed to lock all your keys inside the boxes! Luckily, your home security
system recorded everything, so you know exactly which keys (if any) are
inside each box. You need to get all the keys back out of the boxes, because
they are made of gold. Clearly you have to smash at least one box.

(a) Your baby brother has found the hammer and is eagerly eyeing one
of the boxes. Describe and analyze an algorithm to determine if it is
possible to retrieve all the keys without smashing any box except the
one your brother has chosen.

(b) Describe and analyze an algorithm to compute the minimum number of
boxes that must be smashed to retrieve all the keys.

g. Suppose you are teaching an algorithms course. In your second midterm,
you give your students a drawing of a graph and ask then to indicate a
breadth-first search tree and a depth-first search tree rooted at a particular
vertex. Unfortunately, once you start grading the exam, you realize that the
graph you gave the students has several such spanning trees—far too many

246

Exercises

to list. Instead, you need a way to tell whether each student’s submission is
correct!

In each of the following problems, suppose you are given a connected
graph G, a start vertex s, and a spanning tree T of G.

(a) Suppose G is undirected. Describe and analyze an algorithm to decide
whether T is a depth-first spanning tree rooted at s.

(b) Suppose G is undirected. Describe and analyze an algorithm to decide
whether T is a breadth- first spanning tree rooted at s. [Hint: It’s not
enough for T to be an unweighted shortest-path tree. Yes, this is the
right chapter for this problem!]

(c) Suppose G is directed. Describe and analyze an algorithm to decide
whether T is a breadth- first spanning tree rooted at s. [Hint: Solve
part (b) first.]

(d) Suppose G is directed. Describe and analyze an algorithm to decide
whether T is a depth- first spanning tree rooted at s.

to. Several modern programming languages, including JavaScript, Python,
Perl, and Ruby, include a feature called parallel assignment, which allows
multiple assignment operations to be encoded in a single line of code. For
example, the Python code x, y = 0,1 simultaneously sets x to 0 and y to 1.
The values of the right-hand side of the assignment are all determined by
the old values of the variables. Thus, the Python code a, b = b, a swaps the
values of a and b, and the following Python code computes the nth Fibonacci
number:

def fib(n):

prev, curr =1,0
while n > 0:

prev, curr, n = curr, prev+curr, n-1
return curr

Suppose the interpreter you are writing needs to convert every parallel
assignment into an equivalent sequence of individual assignments. For
example, the parallel assignment a, b = 0,1 can be serialized in either order—
either a=0; b=1 or a=0; b=1 —but the parallel assignment x,y = x+1 ,x+y
can only be serialized as y=x+y; x=x+1 . Serialization may require one or
more additional temporary variables; for example, serializing a, b = b, a
requires one temporary variable, and serializing x, y = x+y, x-y requires two
temporary variables.

(a) Describe an algorithm to determine whether a given parallel assignment
can be serialized without additional temporary variables.

(b) Describe an algorithm to determine whether a given parallel assignment
can be serialized with exactly one additional temporary variable.

247

6. Depth-First Search

Assume that the given parallel assignment involves only simple integer
variables (no indirection via pointers or arrays); no variable appears on the
left side more than once; and expressions on the right side have no side
effects. Don’t worry about the details of parsing the assignment statement;
just assume (but describe!) an appropriate graph representation.

Dynamic Programming

n. Suppose we are given a directed acyclic graph G whose nodes represent jobs
and whose edges represent precedence constraints; that is. each edge u->v
indicates the job u must be completed before job v begins. Each node v also
has a weight T(v) indicating the time required to execute job v.

(a) Describe an algorithm to determine the shortest interval of time in which
all jobs in G can be executed.

(b) Suppose the first job starts at time 0 . Describe an algorithm to determine,
for each vertex v, the earliest time when job v can begin.

(c) Now describe an algorithm to determine, for each vertex v, the latest time
when job v can begin without violating the precedence constraints or
increasing the overall completion time (computed in part (a)), assuming
that every job except v starts at its earliest start time (computed in
part (b)).

12. Let G be a directed acyclic graph with a unique source s and a unique sink t.

(a) A Hamiltonian path in G is a directed path in G that contains every vertex
in G. Describe an algorithm to determine whether G has a Hamiltonian
path.

(b) Suppose the vertices of G have weights. Describe an efficient algorithm
to find the path from s to t with maximum total weight.

(c) Suppose we are also given an integer l . Describe an efficient algorithm
to find the maximum-weight path from s to t that contains at most l
edges. (Assume there is at least one such path.)

(d) Suppose some of the vertices of G are marked as important, and we
are also given an integer k. Describe an efficient algorithm to find the
maximum-weight path from s to t that visits at least k important vertices.
(Assume there is at least one such path.)

(e) Describe an algorithm to compute the number of paths from s to t in G.
(Assume that you can add arbitrarily large integers in 0 ( 1 ) time.)

13. Let G be a directed acyclic graph whose vertices have labels from some fixed
alphabet, and let A[ 1 .. t \ be a string over the same alphabet. Any directed

248

Exercises

path in G has a label, which is a string obtained by concatenating the labels
of its vertices.

(a) Describe an algorithm that either finds a path in G whose label is A or
correctly reports that there is no such path.

(b) Describe an algorithm to find the number of paths in G whose label is A.
(Assume that you can add arbitrarily large integers in 0(1) time.)

(c) Describe an algorithm to find the longest path in G whose label is a
subsequence of A.

(d) Describe an algorithm to find the shortest path in G whose label is a
supersequence of A.

(e) Describe an algorithm to find a path in G whose label has minimum edit
distance from A.

14 . A polygonal path is a sequence of line segments joined end-to-end; the
endpoints of these line segments are called the vertices of the path. The
length of a polygonal path is the sum of the lengths of its segments. A
polygonal path with vertices (x 1; y 1 ), ( x 2 , y 2 ),..., (x k , y k ) is monotonically
increasing if x t < x i+1 and jq < y i+1 for every index i —informally, each
vertex of the path is above and to the right of its predecessor.

Figure 6.20. A monotonically increasing polygonal path with seven vertices through a set of points

Suppose you are given a set S of n points in the plane, represented as
two arrays X[l .. n] and Y[ 1.. n]. Describe and analyze an algorithm to
compute the length of the maximum-length monotonically increasing path
with vertices in S. Assume you have a subroutine Length(x,j, x', y') that
returns the length of the segment from (x, y ) to (x 7 , y').

15 . For any two nodes u and v in a directed acyclic graph G, the interval G[u, v ]
is the union of all directed paths in G from u to v. Equivalently, G[u, v]
consists of all vertices x such that x € reach{u ) and v € reach(x), together
with all the edges in G connecting those vertices.

Suppose we are given a directed acyclic graph G, in which every edge
has a numerical weight, which may be positive, negative, or zero. Describe

249

6. Depth-First Search

an efficient algorithm to find the maximum-weight interval in G, where the
weight of any interval is the sum of the weights of its vertices.

16 . Let G be a directed acyclic graph whose vertices have labels from some fixed
alphabet. Any directed path in G has a label, which is a string obtained by
concatenating the labels of its vertices. Recall that a palindrome is a string
that is equal to its reversal.

(a) Describe and analyze an algorithm to find the length of the longest
palindrome that is the label of a path in G. For example, given the graph
in Figure 6 . 21 , your algorithm should return the integer 6 , which is the
length of the palindrome HANNAH.

(b) Describe an algorithm to find the longest palindrome that is a sub¬
sequence of the label of a path in G.

(c) Describe an algorithm to find the shortest palindrome that is a super¬
sequence of the label of a path in G.

17 . Suppose you are given two directed acyclic graphs G and H in which every
node has a label from some finite alphabet; different nodes may have the
same label. The label of a path in either dag is the string obtained by
concatenating the labels of its vertices.

(a) Describe and analyze an algorithm to compute the length of the longest

string that is both the label of a path in G and the label of a path in H.

(b) Describe and analyze an algorithm to compute the length of the longest

string that is both a subsequence of the label of a path in G both a

subsequence of the label of a path in H.

(c) Describe and analyze an algorithm to compute the length of the shortest
string that is both a supersequence of the label of a path in G both a
supersequence of the label of a path in H.

250

Exercises

18 . Let G be an arbitrary (not necessarily acyclic) directed graph in which every
vertex v has an integer weight w(v).

(a) Describe an algorithm to find the longest directed path in G whose vertex
weights define an increasing sequence.

(b) Describe and analyze an algorithm to determine the maximum-weight
vertex reachable from each vertex in G. That is, for each vertex v, your
algorithm needs to compute max{w(x) | x € reach(v)}.

19 . Suppose you are given a directed graph G in which every edge has negative
weight, and a source vertex s. Describe and analyze an efficient algorithm
that computes the shortest-path distances from s to every other vertex in G.
Specifically, for every vertex t:

• If t is not reachable from s, your algorithm should report dist(t) = 00 .

• If G has a cycle that is reachable from s, and t is reachable from that cycle,
then the shortest-path distance from s to t is not well-defined, because
there are paths (formally, walks) from s to t of arbitrarily large negative
length. In this case, your algorithm should report dist(t) = — 00 .

• If neither of the two previous conditions applies, your algorithm should
report the correct shortest-path distance from s to t.

[Hint: This problem may be easier after you’ve read about shortest paths in
Chapter 8. First think about graphs where the first two conditions never
happen.]

20 . (a) Suppose you are given a directed acyclic graph G with n vertices and an

integer k< n. Describe an efficient algorithm to find a set of at most k
vertex-disjoint paths that visit every vertex in G.

(b) Now suppose the edges of the input dag G have weights, which may be
positive, negative, or zero. Describe an efficient algorithm to find a set
of at most k vertex-disjoint paths with minimum total weight that visit
every vertex in G.

Your algorithms should run in 0(n ,c+c ) time for some small constant c.
A single vertex is a path with weight zero. (We will see a more efficient
algorithm for part (a) in Chapter 11 .)

21 . Kris is a professional rock climber who is competing in the U.S. climbing
nationals. The competition requires Kris to use as many holds on the
climbing wall as possible, using only transitions that have been explicitly
allowed by the route-setter.

The climbing wall has n holds. Kris is given a list of m pairs (x,y) of
holds, each indicating that moving directly from hold x to hold y is allowed;

251

6. Depth-First Search

however, moving directly from y to x is not allowed unless the list also
includes the pair (y, x). Kris needs to figure out a sequence of allowed
transitions that uses as many holds as possible, since each new hold increases
his score by one point. The rules allow Kris to choose the first and last hold
in his climbing route. The rules also allow him to use each hold as many
times as he likes; however, only the first use of each hold increases Kris’s
score.

(a) Define the natural graph representing the input. Describe and analyze
an algorithm to solve Kris’s climbing problem if you are guaranteed that
the input graph is a dag.

(b) Describe and analyze an algorithm to solve Kris’s climbing problem with
no restrictions on the input graph.

Both of your algorithms should output the maximum possible score that Kris
can earn.

22 . There are n galaxies connected by m intergalactic teleport-ways. Each
teleport-way joins two galaxies and can be traversed in both directions.
However, the company that runs the teleport-ways has established an
extremely lucrative cost structure: Anyone can teleport further from their
home galaxy at no cost whatsoever, but teleporting toward their home galaxy
is prohibitively expensive.

Judy has decided to take a sabbatical tour of the universe by visiting as
many galaxies as possible, starting at her home galaxy. To save on travel
expenses, she wants to teleport away from her home galaxy at every step,
except for the very last teleport home.

(a) Describe and analyze an algorithm to compute the maximum number of
galaxies that Judy can visit. Your input consists of an undirected graph
G with n vertices and m edges describing the teleport-way network, an
integer 1 < s < n identifying Judy’s home galaxy, and an array D[1.. n]
containing the distances of each galaxy from s.
r (b) Just before embarking on her universal tour, Judy wins the space lottery,
giving her just enough money to afford two teleports toward her home
galaxy. Describe a new algorithm to compute the maximum number of
distinct galaxies Judy can visit. She can visit the same galaxy more than
once, but crucially, only the first visit counts toward her total.

23 . The Doctor and River Song decide to play a game on a directed acyclic
graph G, which has one source s and one sink t . 8

8 The labels s and t are abbreviations for the Untempered Schism and the Time Vortex, or
the Shining World of the Seven Systems (also known as Gallifrey) and Trenzalore, or Skaro and
Telos, or Something else Timey-wimey. It’s all very complicated, never mind.

252

Exercises

Each player has a token on one of the vertices of G. At the start of the
game, The Doctor’s token is on the source vertex s, and River’s token is on
the sink vertex t. The players alternate turns, with The Doctor moving first.
On each of his turns, the Doctor moves his token forward along a directed
edge; on each of her turns, River moves her token backward along a directed
edge.

If the two tokens ever meet on the same vertex, River wins the game.
(“Hello, Sweetie!”) If the Doctor’s token reaches t or River’s token reaches s
before the two tokens meet, then the Doctor wins the game.

Describe and analyze an algorithm to determine who wins this game,
assuming both players play perfectly. That is, if the Doctor can win no
matter how River moves, then your algorithm should output “Doctor”, and
if River can win no matter how the Doctor moves, your algorithm should
output “River”. (Why are these the only two possibilities?) The input to
your algorithm is the graph G.

* r 2 4 . Let x = x^ 2 ... x n be a given n-character string over some finite alphabet E,
and let A be a deterministic finite-state machine with m states over the same
alphabet.

(a) Describe and analyze an algorithm to compute the length of the longest
subsequence of x that is accepted by A. For example, if A accepts the
language (AR)* and x = ABRACADABRA, your algorithm should output the
number 4 , which is the length of the subsequence ARAR.

(b) Describe and analyze an algorithm to compute the length of the short¬
est supersequence of x that is accepted by A. For example, if A ac¬
cepts the language (ABCDR)* and x = ABRACADABRA, your algorithm
should output the number 25 , which is the length of the supersequence
ABCDRABCDRABCDRABCDRABCDR.

Analyze your algorithms in terms of the length n of the input string, the
number m of states in the finite-state machine, and the size of the alphabet E.

25 . Not every dynamic programming algorithm can be modeled as finding
an optimal path through a directed acyclic graph, but every dynamic
programming algorithm does process some underlying dependency graph
in postorder.

(a) Suppose we are given a directed acyclic graph G where every node stores
a numerical search key. Describe and analyze an algorithm to find the
largest binary search tree that is a subgraph of G.

(b) Suppose we are given a directed acyclic graph G and two vertices s and t.
Describe an algorithm to compute the number of directed paths in G
from s to t. (Assume that any arithmetic operation requires 0(1) time.)

253

6. Depth-First Search

(c) Let G be a directed acyclic graph with the following features:

• G has a single source s and several sinks t l3 t 2 ,..., t^.

• Each edge v->w has an associated weight p(v->w ) between 0 and 1.

• For each non-sink vertex v, the total weight of all edges leaving v
is 1; that is, £ w p(v^w) = L

The weights p(v^w) define a random walk in G from the source s to some
sink tp, after reaching any non-sink vertex v, the walk follows edge v->w
with probability p(v-nv). All probabilities are mutually independent.
Describe and analyze an algorithm to compute the probability that this
random walk reaches sink t ; , for every index i. (Assume that each
arithmetic operation takes only 0(1) time.)

254

We must all hang together, gentlemen,

or else we shall most assuredly hang separately.

— Benjamin Franklin, at the signing of the
Declaration of Independence (July 4,1776)

I remember seeking advice from someone—who could it have been?—about
whether this work was worth submitting for publication; the reasoning it uses is so
very simple.... Fortunately he advised me to go ahead, and many years passed
before another of my publications became as well-known as this very simple one.

— Joseph Kruskal, describing his shortest-spanning-subtree algorithm (1997)

Clean ALL the things!

— Allie Brosh, "This is Why I'll Never be an Adult",
Hyperbole and a Half, June 17, 2010.

7

Minimum Spanning Trees

Suppose we are given a connected, undirected, weighted graph. This is a
graph G = (V, E) together with a function w: E —» R that assigns a real weight
w(e) to each edge e, which may be positive, negative, or zero. This chapter
describes several algorithms to find the minimum spanning tree of G, that is,
the spanning tree T that minimizes the function

w(T) :=^]w(e).

eeT

See Figure 7.1 for an example.

7.1 Distinct Edge Weights

An annoying subtlety in the problem statement is that weighted graphs can
have more than one spanning tree with the same minimum weight; in particular,

255

7. Minimum Spanning Trees

Figure 7.1 . A weighted graph and its minimum spanning tree.

if every edge in G has weight 1, then every spanning tree of G is a minimum
spanning tree, with weight V — 1. This ambiguity complicates the development
of our algorithms; everything would be much simpler if we could simply assume
that minimum spanning trees are unique.

Fortunately, there is an easy condition that implies the uniqueness we want.

Lemma 7 . 1 . If all edge weights in a connected graph G are distinct, then G has
a unique minimum spanning tree. 1

Proof: Let G be an arbitrary connected graph with two minimum spanning
trees T and T'; we need to prove that some pair of edges in G have the same
weight. The proof is essentially a greedy exchange argument.

Each of our spanning trees must contain an edge that the other tree omits.
Let e be a minimum-weight edge in T \ T', and let e' be a minimum-weight
edge in T' \ T (breaking ties arbitrarily). Without loss of generality, suppose
w(e) < w(e').

The subgraph T' U {e} contains exactly one cycle C, which passes through
the the edge e. Let e" be any edge of this cycle that is not in T. At least one
such edge must exist, because T is a tree. (We may or may not have e" = e'I)
Because e e T, we immediately have e" 7 ^ e and therefore e" e T' \T. It follows
that w(e") > w{e') > w(e).

Now consider the spanning tree T" = T' + e — e". (This new tree T" might
be equal to T.) We immediately have w(T ") = w(T') + w(e) — w(e") < w(T').
But T' is a minimum spanning tree, so we must have w(T // ) = w(T'); in other
words, T" is also a minimum spanning tree. We conclude that w(e) = w(e"),
which completes the proof. □

If we already have an algorithm that assumes distinct edge weights, we can
still run it on graphs where some edges have equal weights, as long as we have

‘The converse of this lemma is false; a connected graph with repeated edge weights can still
have a unique minimum spanning tree. As a trivial example, suppose G is a tree!

256

7.2. The Only Minimum Spanning Tree Algorithm

a consistent method for breaking ties. One such method uses the following
algorithm in place of simple weight comparisons. ShorterEdge takes as input
four integers i,j,k,l, representing four (not necessarily distinct) vertices, and
decides which of the two edges (i,;) and (k, l ) has “smaller” weight. (Because
the input graph undirected, the pairs (i,j) and (j, i) represent the same edge.)

ShorterEdge(i, ;,k, Z)
if w(i,j ) < w(k, Z)
if w(i,j ) > w(k, Z)
if min(i, j) < min(k, Z)
if min(i, j) > min(k, Z)
if max(i, j) < max(Zc, Z)
((ifmax(ij) > max(k,l )))

then return (i, j)
then return (k, Z)
then return (i, j)
then return (k, Z)
then return (i, j)
return (k, Z)

In light of Lemma 7.1 and this tie-breaking rule, we will safely assume for
the rest of this chapter that edge weights are always distinct, and therefore
minimum spanning trees are always unique. In particular, we can freely discuss
the minimum spanning tree with no confusion.

7.2 The Only Minimum Spanning Tree Algorithm

There are many algorithms to compute minimum spanning trees, but almost all
of them are instances of the following generic strategy. The situation is similar
to graph traversal, where several different algorithms are all variants of the
generic traversal algorithm whatever-first search.

The generic minimum spanning tree algorithm maintains an acyclic sub¬
graph F of the input graph G, which we will call the intermediate spanning forest.
At all times, F satisfies two invariants:

• F is a subgraph of the minimum spanning tree of G.

• Every component of F is a minimum spanning tree of its vertices.

(In fact, the second invariant implies the first.) Initially, F consists of V one-node
trees. The generic algorithm merges trees in F together by adding certain
edges between them. When the algorithm halts, F consists of a single spanning
tree; our invariants imply that this must be the minimum spanning tree of G.
Obviously, we have to be careful about which edges we add to the evolving
forest, since not every edge is in the minimum spanning tree.

At any stage of its evolution, the intermediate spanning forest F induces
two special types of edges.

• An edge is useless if it is not an edge of F, but both its endpoints are in the
same component of F.

• An edge is safe if it is the minimum-weight edge with with exactly one
endpoint in some component of F.

257

7. Minimum Spanning Trees

The same edge could be safe for two different components off. Some edges of
G\F are neither safe nor useless; we call these edges undecided.

All minimum spanning tree algorithms are based on two simple observations.
The first observation was proved by Robert Prim in t957 (although it is implicit
in several earlier algorithms), and the second is immediate.

Lemma 7.2 (Prim). The minimum spanning tree of G contains every safe edge.

Proof: In fact we prove the following stronger statement: For any subset S of
the vertices of G, the minimum spanning tree of G contains the minimum-weight
edge with exactly one endpoint in S. Like the previous lemma, we prove this
claim using a greedy exchange argument.

Let S be an arbitrary subset of vertices of G, and let e be the lightest edge
with exactly one endpoint in S. (Our assumption that all edge weights are
distinct implies that e is unique.) Let T be an arbitrary spanning tree that does
not contain e; we need to prove that T is not the minimum spanning tree of G.

Because T is connected, it contains a path from one endpoint of e to the
other. Because this path starts at a vertex of S and ends at a vertex not in S, it
must contain at least one edge with exactly one endpoint in S; let e' be any such
edge. Because T is acyclic, removing e' from T yields a spanning/orest with
exactly two components, one containing each endpoint of e. Thus, adding e
to this forest gives us a new spanning tree T' = T — e' + e. The definition of e
implies w(e') > w(e), which implies that T' has smaller total weight than T.
Thus, T is not the minimum spanning tree of G, which completes the proof. □

Figure 7.2. Every safe edge is in the minimum spanning tree. Black vertices are in the subset S.

Lemma 7.3. The minimum spanning tree contains no useless edge.

Proof: Adding any useless edge to F would introduce a cycle. □

Our generic minimum spanning tree algorithm repeatedly adds safe edges to
the evolving forest F. Whenever we add new edges to F, some undecided edges
become safe, and others become useless. (Once an edge becomes useless, it
stays useless forever.) To specify a particular algorithm, we must specify which
safe edge(s) to add in each iteration, and we must describe how to find those
edges.

258

73- Boruvka's Algorithm

7.3 Boruvka's Algorithm

The oldest and arguably simplest minimum spanning tree algorithm was discov¬
ered by the Czech mathematician Otakar Boruvka in 1926, about a year after
Jindrich Saxel asked him how to construct an electrical network connecting
several cities using the least amount of wire. 2 The algorithm was rediscovered by
Gustav Choquet in 1938, rediscovered again by a team of Polish mathematicians
led by Jozef Lukaszewicz in 1951, and rediscovered again by George Sollin in
1961. Although Sollin never published his rediscovery, it was carefully described
and credited in one of the first textbooks on graph algorithms; as a result, this
algorithm is sometimes called “Sollin’s algorithm”.

The Boruvka / Choquet / Florek-Lukaziewicz-Perkal-Steinhaus-Zubrzycki/
Prim/Sollin/Brosh 3 algorithm can be summarized in one line:

Boruvka: Add ALL the safe edges and recurse.

Figure 7.3. Boruvka's algorithm run on the example graph. Thick red edges are in F; dashed edges are
useless. Arrows point along each component's safe edge. The algorithm ends after just two iterations.

Here is Boruvka’s algorithm in more detail. The algorithm calls the Count-
AndLabel algorithm from Chapter 5 (on page 202) to count the components
of F and label each vertex v with an integer comp(v) indicating its component.

Boruvka(V,E):

F = (Y, 0 )

count <— CountAndLabel(F)
while count > 1

AddAllSafeEdges(£, F, count )
count <— CountAndLabel(F)
return F

2 Saxel was an employee of the West Moravian Power Company, described by Boruvka as
“very talented and hard-working”, who was later executed by the Nazis as a person of Jewish
descent.

3 Go read everything in Hyperbole and a Half. And then go buy the book. And an extra copy
for your cat. What’s that? You don’t have a cat? What kind of a monster are you? Go get a cat,
and then buy it an extra copy of Hyperbole and a Half.

259

7. Minimum Spanning Trees

It remains only to describe how to identify and add all the safe edges to F.
Suppose F has more than one component, since otherwise we’re already done.
The following subroutine computes an array safe[l .. V] of safe edges, where
safe[i] is the minimum-weight edge with one endpoint in the ith component
of F, by a brute force examination of every edge in G. For each edge uv, if u
and v are in the same component, then uv is either useless or already an edge
in F. Otherwise, we compare the weight of uv to the weights of safe[comp(u)]
and safe[comp(v )] and update the array entries if necessary. Once we have
identified all the safe edges, we add each edge safe[i] to F.

AddAllSafeEdges(E, F, count ) :
for i «— 1 to count
safe[i ] <— Null
for each edge uv e£

if comp{ii) ^ comp{y )

if safe[comp{u)] = Null or w(uv) < w(safe[comp(u)])
safe[comp(u )] <— uv

if safe[comp(v)] = Null or w(uv) < w(safe[comp(y)])
sa/e[ZabeZ(v)] <— uv
for i <— 1 to count
add sa/e[i] to F

Each call to CountAndLabel runs in O(V) time, because the forest F has
at most V — 1 edges. AddAllSafeEdges runs in 0 {V + E) time, because we spend
constant time on each vertex, each edge of G, and each component of F. Because
the input graph is connected, we have V < E + 1 . It follows that each iteration
of the while loop of Boruvka takes 0 (E) time.

Each iteration reduces the number of components of F by at least a factor
of two—in the worst case, the components of F coalesce in pairs. Because F
initially has V components, the while loop iterates at most O(logV) times. We
conclude that the overall running time of Boruvka’s algorithm is 0 (E log V).

This is the MST Algorithm You Want

Despite its relatively obscure origin, early algorithms researchers were aware
of Boruvka’s algorithm, but dismissed it as being “too complicated”. As a
result, despite its simplicity and efficiency, most algorithms and data structures
textbooks unfortunately never even mention Boruvka’s algorithm. This omission
is a serious mistake; Boruvka’s algorithm has several distinct advantages over
other classical MST algorithms.

• Boruvka’s algorithm often runs faster than the 0 {E log V ) worst-case running
time. The number of components in F can drop by significantly more than a
factor of 2 in a single iteration, reducing the number of iterations below the
worst-case [log 2 V].

260

7 - 4 - Jarnfk's ("Prim's") Algorithm

• A slight reformulation of Boruvka’s algorithm (actually closer to Boruvka’s
original presentation) actually runs in 0(E) time for a broad class of
interesting graphs, including graphs that can be drawn in the plane without
edge crossings. In contrast, the time analysis for the other two algorithms
applies to all graphs.

• Boruvka’s algorithm allows for significant parallelism; in each iteration,
each component of F can be handled in a separate independent thread.
This implicit parallelism allows for even faster performance on multicore or
distributed systems. In contrast, the other two classical MST algorithms are
intrinsically serial.

• There are several more recent minimum-spanning-tree algorithms that are
faster even in the worst case than the classical algorithms described here.
All of these faster algorithms are generalizations of Boruvka’s algorithm.

In short, if you ever need to implement a minimum-spanning-tree algorithm,
use Boruvka. On the other hand, if you want to prove things about minimum
spanning trees effectively, you really need to know the next two algorithms as
well.

7.4 Jarnfk's ("Prim's") Algorithm

The next oldest minimum spanning tree algorithm was first described by the
Czech mathematician Vojtech Jarnik in a t929 letter to Boruvka; Jarnik published
his discovery the following year. The algorithm was independently rediscovered
by Joseph Kruskal in 1956, (arguably) by Robert Prim in 1957, and finally by
Edsger Dijkstra in 1958. Both Prim and Dijkstra (eventually) knew of and even
cited Kruskal’s paper, but since Kruskal also described two other minimum-
spanning-tree algorithms in the same paper, this algorithm is usually called
“Prim’s algorithm”, or sometimes “the Prim/Dijkstra algorithm”, even though by
1958 Dijkstra already had another algorithm (inappropriately) named after him.

In Jarnfk’s algorithm, the intermediate forest F has only one nontrivial
component T ; all the other components are isolated vertices. Initially, T consists
of a single arbitrary vertex of the graph. The algorithm repeats the following
step until T spans the whole graph:

Jarnik: Repeatedly add 7 ” s safe edge to T.

To implement Jarnfk’s algorithm, we keep all the edges adjacent to T in
a priority queue. When we pull the minimum-weight edge out of the priority
queue, we first check whether both of its endpoints are in T. If not, we add the
edge to T and then add the new neighboring edges to the priority queue. In
other words, Jarnfk’s algorithm is a variant of “best first search”, as described at

261

7. Minimum Spanning Trees

Figure 7.4. Jarnik's algorithm run on the example graph, starting with the bottom vertex. At each stage,
thick red edges are in T, an arrow points along T's safe edge; and dashed edges are useless.

the end of Chapter 5! If we implement the underlying priority queue using a
standard binary heap, Jarnik’s algorithm runs in 0 {E log£) = 0(E log V) time.

^Improving Jarnik's Algorithm

We can improve Jarnik’s algorithm using a more complex priority queue data
structure called a Fibonacci heap, first described by Michael Fredman and
Robert Tarjan in 1984. Just like binary heaps, Fibonacci heaps support the
standard priority queue operations Insert, ExtractMin, and DecreaseKey.
However, unlike standard binary heaps, which require 0 (log n) time for every
operation, Fibonacci heaps support Insert and DecreaseKey in constant
amortized time. The amortized cost of ExtractMin is still O(logn). 4

To apply this faster data structure, we keep the vertices of G in the priority
queue instead of edges, where the priority of each vertex v is either the minimum-
weight edge between v and the evolving tree T, or 00 if there is no such edge.
We can Insert all the vertices into the priority queue at the beginning of the
algorithm; then, whenever we add a new edge to T, we may need to decrease
the priorities of some neighboring vertices.

To make the description easier, we break the algorithm into two parts.
JarnikInit initializes the priority queue; JarnikLoop is the main algorithm.

4 Amortized time is an accounting trick that allows us to ignore infrequent fluctuations in
the time for a single data structure operation. A Fibonacci heap can execute any intermixed
sequence of I Inserts, D DecreaseKeys, and X ExtractMins in 0(7 + D + X log n) time, in the
worst case. So the average Insert and the average DecreaseKey each take constant time, and
the average ExtractMin takes O(logn) time; however, some individual operations may take
longer in the worst case. Amortization uses statistical averaging over the sequence of operations;
there is no assumption of randomness here, either in the input data or in the algorithm.

262

7 - 5 - Kruskal's Algorithm

The input consists of the vertices and edges of the graph, along with the start
vertex s. For each vertex v, we maintain both its priority priority(y) and the
incident edge edge(v) such that w(edge(v)) = priority(v).

Jarnik(V, E,s):
JarnikInit(V, E,s )
JarnikLoop( V, E, s )

JarnikInit(V,£,s):

JarnikLoop( V,E,s ):

for each vertex veV\{s[

T^C{s},0)

if vs e E

for i <— 1 to |Vj — 1

edge(v) <— vs

v <— ExtractMin

priorityiy ) <— w(vs)

add v and edge(v) to T

else

for each neighbor u of v

edge(v) <— Null

if u <£. T and priority(u ) > w(izv)

priority(v) <— 00

edge(u) <— uv

Insert(v)

DecreaseKey(u, w(uv))

Figure 7.5. Jarnik's minimum spanning tree algorithm, ready to be used with a Fibonacci heap

The operations Insert and ExtractMin are each called O(V) times once
for each vertex except s, and DecreaseKey is called 0 (E) times, at most twice
for each edge. Thus, if we use a Fibonacci heap, the improved algorithm runs in
0(E + Vlog V) time, which is faster than Boruvka’s algorithm unless E = 0(V).

In practice, however, this improvement is rarely faster than the naive
implementation using a binary heap, unless the graph is extremely large and
dense. The Fibonacci heap algorithms are quite complex, and the hidden
constants in both the running time and space are significant—not outrageous,
but certainly bigger than the hidden constant 1 in the O(logn) time bound for
binary heap operations.

7.5 Kruskal's Algorithm

The last minimum spanning tree algorithm we’ll consider was first described by
Joseph Kruskal in 1956, in the same paper where he rediscovered Jarnik’s algo¬
rithm. Kruskal was motivated by “a typewritten translation (of obscure origin)”
of Boruvka’s original paper that had been “floating around” the Princeton math
department. Kruskal found Boruvka’s algorithm “unnecessarily elaborate”. 5
The same algorithm was rediscovered in 1957 by Loberman and Weinberger, but
somehow avoided being renamed after them.

5 To be fair, Boruvka’s first paper was unnecessarily elaborate, in part because it was written
for mathematicians in the formal language of (linear) algebra, rather than in the language of
graphs. Boruvka’s followup paper, also published in 1927 but in an electrotechnical journal, was
written in plain language for a much broader audience, essentially its current modern form.
Kruskal was apparently unaware of Boruvka’s second paper. Stupid Iron Curtain.

263

7. Minimum Spanning Trees

Like our earlier minimum-spanning tree algorithms, Kruskal’s algorithm has
a memorable one-line description:

Kruskal: Scan all edges by increasing weight; if an edge is safe, add it to F.

Figure 7.6. Kruskal's algorithm run on the example graph. Thick red edges are in F, thin dashed edges
are useless.

The simplest method to scan the edges in increases weight order is to sort
the edges by weight, in 0(Elog£) time, and then use a simple for-loop over the
sorted edge list. As we will see shortly, this preliminary sorting dominates the
running time of the algorithm.

Because we examine the edges in order from lightest to heaviest, any edge
we examine is safe if and only if its endpoints are in different components of the
forest F. Suppose we encounter an edge e joins two components A and B but is
not safe. Then there mut be a lighter edge e' with exactly one endpoint in A.
But this is impossible, because (inductively) every previously examined edge
has both endpoints in the same component of F.

Just as in Boruvka’s algorithm, each vertex of F needs to "know" which
component of F contains it. Unlike Boruvka’s algorithm, however, we do
not recompute all component labels from scratch every time we add an edge.
Instead, when two components are joined by an edge, the smaller component
inherits the label of the larger component; that is, we traverse the smaller
component (via whatever-first search). This traversal requires 0 ( 1 ) time for
each vertex in the smaller component. Each time the component label of a
vertex changes, the component of F containing that vertex grows by at least a
factor of 2 ; thus, each vertex label changes at most 0 (log V) times. It follows
that the total time spent updating vertex labels is only 0(V log V).

264

Exercises

More generally, Kruskal’s algorithm maintains a partition of the the vertices
of G into disjoint subsets (in our case, the components of F), using a data
structure that supports the following operations:

• MakeSet(v) — Create a set containing only the vertex v.

• Find(v) — Return an identifier unique to the set containing v.

• Union(u, v) — Replace the sets containing u and v with their union. (This
operation decreases the number of sets.)

Here’s a complete description of Kruskal’s algorithm in terms of these operations:

Kruskal(V, F):

sort E by increasing weight
F <— (V, 0)
for each vertex veF
MakeSet(v)
for i <— 1 to |F|

uv <— ith lightest edge in E
if Find(u) Find(v)

Union(u, v)
add uv to F

return F

After the initial sort, the algorithm performs exactly V MakeSet operations
(one for each vertex), 2 E Find operations (two for each edge), and V — 1 Union
operations (one for each edge in the minimum spanning tree). We just described
a disjoint-set data structure for which MakeSet and Find require 0(1) time,
and Union runs in O(logV) amortized time. Using this implementation, the
total time spent maintaining the set partition is 0(F + V logU). 6

But recall that we already need O(FlogF) = O(FlogV) time just to sort the
edges. Because this is larger than the time spent maintaining the Union-Find
data structure, the overall running time of Kruskal’s algorithm is O(FlogV),
exactly the same as Boruvka’s algorithm, or Jarnik’s algorithm with a normal
(non-Fibonacci) heap.

Exercises

i. Let G = (V, F) be an arbitrary connected graph with weighted edges.

(a) Prove that for any cycle in G, the minimum spanning tree of G excludes
the maximum-weight edge in that cycle.

6 A different disjoint-set data structure, which uses a strategy called union-by-rank with
path compression, performs each Union or Find in 0(a(V)) amortized time, where a is the
almost-but-not-quite-constant inverse-Ackerman function. If you don’t feel like using Wikipedia,
just think of a(V) as 4. Using this implementation, the total time spent maintaining the set
partition is 0(Ea(V)), which is slightly faster when V is large and E is very close to V.

265

7. Minimum Spanning Trees

(b) Prove or disprove: The minimum spanning tree of G includes the
minimum-weight edge in every cycle in G.

2. Throughout this chapter, we assumed that no two edges in the input graph
have equal weights, which implies that the minimum spanning tree is unique.
In fact, a weaker condition on the edge weights implies MST uniqueness.

(a) Describe an edge-weighted graph that has a unique minimum spanning
tree, even though two edges have equal weights.

(b) Prove that an edge-weighted graph G has a unique minimum spanning
tree if and only if the following conditions hold:

• For any partition of the vertices of G into two subsets, the minimum-
weight edge with one endpoint in each subset is unique.

• The maximum-weight edge in any cycle of G is unique.

(c) Describe and analyze an algorithm to determine whether or not a graph
has a unique minimum spanning tree.

3. Most classical minimum-spanning-tree algorithms use the notions of “safe”
and “useless” edges described in the text, but there is an alternate formulation.
Let G be a weighted undirected graph, where the edge weights are distinct.
We say that an edge e is dangerous if it is the longest edge in some cycle
in G, and useful if it does not lie in any cycle in G.

(a) Prove that the minimum spanning tree of G contains every useful edge.

(b) Prove that the minimum spanning tree of G does not contain any
dangerous edge.

(c) Describe and analyze an efficient implementation of the following
algorithm, first described by Kruskal in the same 1956 paper where he
proposed “Kruskal’s algorithm”. Examine the edges of G in decreasing
order; if an edge is dangerous, remove it from G. [Hint: It won’t be as
fast as Kruskal’s usual algorithm.]

4. (a) Describe and analyze an algorithm to compute the maximum -weight

spanning tree of a given edge-weighted graph.

(b) A feedback edge set of an undirected graph G is a subset F of the edges
such that every cycle in G contains at least one edge in F. In other
words, removing every edge in F makes the graph G acyclic. Describe
and analyze a fast algorithm to compute the minimum weight feedback
edge set of of a given edge-weighted graph.

5. Suppose we are given both an undirected graph G with weighted edges and
a minimum spanning tree T of G.

266

Exercises

(a) Describe an algorithm to update the minimum spanning tree when the
weight of a single edge e is decreased.

(b) Describe an algorithm to update the minimum spanning tree when the
weight of a single edge e is increased.

In both cases, the input to your algorithm is the edge e and its new weight;
your algorithms should modify T so that it is still a minimum spanning tree.
[Hint: Consider the cases e e T and e ^ T separately.]

6. (a) Describe and analyze an algorithm to find the second smallest spanning

tree of a given graph G, that is, the spanning tree of G with smallest
total weight except for the minimum spanning tree.
r (b) Describe and analyze an efficient algorithm to compute, given a weighted
undirected graph G and an integer k, the k spanning trees of G with
smallest weight.

7. A graph G = (V,E) is dense if E = 0 (V 2 ). Describe a modification of Jarnik’s
minimum-spanning tree algorithm that runs in 0 (V 2 ) time (independent
of E) when the input graph is dense, using only simple data structures (and
in particular, without using a Fibonacci heap).

8. Minimum-spanning tree algorithms are often formulated using an operation
called edge contraction. To contract the edge izv, we insert a new node,
redirect any edge incident to u or v (except uv) to this new node, and then
delete u and v. After contraction, there may be multiple parallel edges
between the new node and other nodes in the graph; we remove all but the
lightest edge between any two nodes.

Figure 7.7. Contracting an edge and removing redundant parallel edges.

The three classical minimum-spanning tree algorithms described in this
chapter can all be expressed cleanly in terms of contraction as follows. All
three algorithms start by making a clean copy G' of the input graph G
and then repeatedly contract safe edges in G'; the minimum spanning tree
consists of the contracted edges.

• Boruvka: Mark the lightest edge leaving each vertex, contract all
marked edges, and recurse.

267

7. Minimum Spanning Trees

• Jarnik: Repeatedly contract the lightest edge incident to some fixed
root vertex.

• Kruskal: Repeatedly contract the lightest edge in the graph.

(a) Describe an algorithm to execute a single pass of Boruvka’s contraction
algorithm in 0 ( 1 / + E) time. The input graph is represented in an
adjacency list.

(b) Consider an algorithm that first performs k passes of Boruvka’s contrac¬
tion algorithm, and then runs Jarnik’s algorithm ( with a Fibonacci heap)
on the resulting contracted graph.

i. What is the running time of this hybrid algorithm, as a function of
V, E, and k ?

ii. For which value of k is this running time minimized? What is the
resulting running time?

(c) Call a family of graphs nice if it has the following properties:

• A nice graph with n vertices has only O(n) edges.

• Contracting an edge of a nice graph yields another nice graph.

For example, planar graphs—graphs that can be drawn in the plane with
no crossing edges—are nice. Euler’s formula implies that any planar
graph with 1 / vertices has at most 31 / — 6 edges, and contracting any
edge of a planar graph leaves a smaller planar graph.

Prove that Boruvka’s contraction algorithm computes the minimum
spanning tree of any nice graph in 0(1/) time.

9. Consider a path between two vertices s and t in a undirected weighted
graph G. The width of this path is the minimum weight of any edge in the
path. The bottleneck distance between s and t is the width of the widest
path from s to t. (If there are no paths from s to t, the bottleneck distance
is —00; on the other hand, the bottleneck distance from s to itself is 00.)

The bottleneck distance between s and t is 9.

(a) Prove that the maximum spanning tree of G contains widest paths
between every pair of vertices.

268

Exercises

(b) Describe an algorithm to solve the following problem in 0 (V + E ) time:
Given a undirected weighted graph G, two vertices s and t, and a
weight W, is the bottleneck distance between s and t at most W?

(c) Suppose B is the bottleneck distance between s and t.

i. Prove that deleting any edge with weight less than B does not change
the bottleneck distance between s and t.

ii. Prove that contracting any edge with weight greater than B does
not change the bottleneck distance between s and t. (If contraction
creates parallel edges, delete all but the heaviest edge between each
pair of nodes.)

(d) Describe an algorithm to compute a minimum-bottleneck path between s
and t in 0 {V + E) time. [Hint: Start by finding the median-weight edge
in G.]

*10. (This problem assumes familiarity with standard implementations of disjoint-
set data structures.)

Consider the following variant of Boruvka’s algorithm. Instead of
counting and labeling components of F to find safe edges, we use a standard
disjoint set data structure. Each component of F is represented by an
up-tree; each vertex v stores a pointer parent{v ) to its parent in the up-tree
containing v. Each leader vertex v also maintains an edge sa/e(v), which is
(eventually) the lightest edge with one endpoint in v’s component of F.

Boruvka( V,E):

F = 0

for each vertex veV
parentfv ) <— v
while FindSafeEdges(V,£)
AddSafeEdges(V,£, F)
return F

FindSafeEdges(V, E ) :
for each vertex v e V
safety) <— Null
found <— False
for each edge uv e E

u <— Find(k); v <— Find(v)
if u ^ v

if w(uv ) < w(safe(u ))
safety) <— uv
if wfuv) < wtsafefv))
safetv) <— uv
found <— True
return found

AddSafeEdges (V,E,E) :
for each vertex v e V
if safetv) ^ Null
xy <— safetv)
if Find(x) / FiND(y)
UNiON(x,y)
add xy to F

269

7. Minimum Spanning Trees

Prove that if Find uses path compression, then each call to FindSafeEdges
and AddSafeEdges requires only 0 {V + E) time. [Hint: It doesn’t matter
how Union is implemented! What is the depth of the up-trees when
FindSafeEdges ends?]

270

I study my Bible as i gather apples. First I shake the whole tree, that the ripest might
fall. Then I climb the tree and shake each limb, and then each branch and then each
twig, and then I look under each leaf.

— attributed to Martin Luther (c. 1500)

Life is an unfoldment, and the further we travel the more truth we can comprehend.
To understand the things that are at our door is the best preparation for
understanding those that lie beyond.

— attributed to Hypatia of Alexandria (c. 400) by Elbert Hubbard
in Little Journeys to the Homes of Great Teachers (1908)

Your mind will answer most questions if you learn to relax and wait for the answer.
Like one of those thinking machines, you feed in your question, sit back, and wait...

- William S. Burroughs. Naked Lunch (1959)

The methods given in this paper require no foresight or ingenuity,
and hence deserve to be called algorithms.

— Edward R. Moore, "The Shortest Path Through a Maze” (1959)

Shortest Paths

Suppose we are given a weighted directed graph G = (V, E, w) with two special
vertices, and we want to find the shortest path from a source vertex s to a target
vertex t. That is, we want to find the directed path P starting at s and ending
at t that minimizes the function

w(P):= Z w(u->v).

u-»veP

For example, if I want to answer the question “What’s the fastest way to drive
from my old apartment in Champaign, Illinois to my wife’s old apartment in
Columbus, Ohio?”, I might use a graph whose vertices are cities, edges are
roads, weights are driving times, s is Champaign, and t is Columbus. 1 The
graph is directed, because driving times along the same road might be different

‘West on Church, north on Prospect, east on I-74, south on I-465, east on Airport Expressway,
north on I-65, east on I-70, north on Grandview, east on 5th, north on Olentangy River, east on
Dodridge, north on High, west on Kelso, south on Neil. Depending on traffic. We live in Urbana
now.

271

8. Shortest Paths

in different directions. (At one time, there was a speed trap on I-70 just east of
the Indiana/Ohio border, but only for eastbound traffic.)

8.1 Shortest Path Trees

Almost every algorithm known for computing shortest paths from one vertex
to another actually solves (large portions of) the following more general single
source shortest path or SSSP problem: Find shortest paths from the source
vertex s to every other vertex in the graph. This problem is usually solved by
finding a shortest path tree rooted at s that contains all the desired shortest
paths.

It’s not hard to see that if shortest paths are unique, then they form a tree,
because any subpath of a shortest path is itself a shortest path. If there are
multiple shortest paths to some vertices, we can always choose one shortest
path to each vertex so that the union of the paths is a tree. If there are shortest
paths to two vertices u and v that diverge, then meet, then diverge again, we
can modify one of the paths without changing its length so that the two paths
only diverge once.

Figure 8.1. If s^a->b->c->d->v (solid) and s^>a^>x^y->d^u (dashed) are shortest paths, then
s^>a^>b^>c^>d^>u (along the top) is also a shortest path.

Although they are both optimal spanning trees, shortest-path trees and
minimum spanning trees are very different creatures. Shortest-path trees are
rooted and directed; minimum spanning trees are unrooted and undirected.
Shortest-path trees are most naturally defined for directed graphs; minimum
spanning trees are more naturally defined for undirected graphs. If edge weights
are distinct, there is only one minimum spanning tree, but every source vertex
induces a different shortest-path tree; moreover, it is possible for every shortest
path tree to use a different set of edges from the minimum spanning tree.

*8.2 Negative Edges

For most shortest-path problems, where the edge weights correspond to distance
or length or time, it is natural to assume that all edge weights are non-negative,
or even positive. However, for many applications of shortest-path algorithms, it
is natural to consider negative edges. For example, the weight of an edge might

272

▼8.2. Negative Edges

Figure 8.2. A minimum spanning tree and a shortest path tree of the same undirected graph.

represent the cost of moving from one vertex to another, so negative-weight
edges represent transitions with negative cost, or equivalently, transitions that
earn a profit.

Negative edges are a thorn in the side of most shortest-path problems,
because the presence of a negative cycle might imply that there is no shortest
path. To be precise, a shortest path from s to t exists if and only if there is at
least one path from s to t, but there is no path from s to t that touches a negative
cycle. For any path from s to t that touches a negative cycle, there is a shorter
path from s to t that goes around the cycle one more time. 2 Thus, if at least one
path from s to t touches a negative cycle, there is no shortest path from s to t.

In part because we need to consider negative wedge weights, this chapter
explicitly considers only directed graphs. All of the algorithms described here
also work for undirected graphs with essentially trivial modifications, if and
only if negative edges are prohibited. Correctly handling negative edges in
undirected graphs is considerably more subtle. We cannot simply replace every
undirected edge with a pair of directed edges, because this would transform any
negative edge into a short negative cycle. Subpaths of an undirected shortest path
that contains a negative edge are not necessarily shortest paths; consequently,
the set of all undirected shortest paths from a single source vertex may not
define a tree, even if shortest paths are unique.

technically, we should be discussing shortest walks here, rather than shortest paths, but the
abuse of terminology is standard. If s can reach t, there must be a shortest simple path from s
to t; it’s just NP-hard to compute (when there are negative cycles). On the other hand, if there is
a shortest walk from s to t, that walk must be a simple path, and therefore must be the shortest
simple path from s to t. Blerg.

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8. Shortest Paths

Figure 8.4. An undirected graph where shortest paths from 5 are unique but do not define a tree.

A complete treatment of undirected graphs with negative edges is beyond
the scope of this book. I will only mention, for people who want to follow up
via Google, that a single shortest path in an undirected graph with negative
edges can be computed in 0 {VE + V 2 log V) time, by a reduction to maximum
weighted matching.

8.3 The Only SSSP Algorithm

Just like graph traversal and minimum spanning trees, many different SSSP
algorithms can be described as special cases of a single generic algorithm, first
proposed by Lester Ford in 1956 and independently described by George Dantzig
in 1957 and again by George Minty in 1958. 3 Each vertex v in the graph stores
two values, which (inductively) describe a tentative shortest path from s to v.

• dist(v ) is the length of the tentative shortest s~>v path, or 00 if there is no
such path.

• pred(v) is the predecessor of v in the tentative shortest s~>v path, or Null
if there is no such vertex.

The predecessor pointers automatically define a tentative shortest-path tree
rooted at s; these pointers are exactly the same as the parent pointers in
our generic graph traversal algorithm. At the beginning of the algorithm, we
initialize the distances and predecessors as follows:

InitSSSP(s):
dist(s) <— 0
pred(s) <— Null
for all vertices vjfs
dist(v) <— 00
pred(v) <— Null

During the execution of the algorithm, an edge u^v is tense it dist(u)+w(u->v) <
dist(v). If u—>v is tense, the tentative shortest path is clearly incorrect,
because the path s~>u->v is shorter. We can correct (or at least improve) this
obvious overestimate by relaxing the edge as follows:

Specifically, Dantzig showed that the shortest path problem can be phrased as a linear
programming problem, and then described an interpretation of his simplex method in terms of
the original graph. His description is (morally) equivalent to Ford’s relaxation strategy.

274

8.3. The Only SSSP Algorithm

Relax(u^v):

dist(v) <— dist(u ) + w(u->v)
pred(v)<— u

Now that everything is set up, Ford’s generic algorithm has a simple one-line
description:

Repeatedly relax tense edges, until there are no more tense edges.

FqrdSSSP(s):

InitSSSP(s)

while there is at least one tense edge
Relax any tense edge

If FordSSSP eventually terminates (because there are no more tense edges),
then the predecessor pointers correctly define a shortest-path tree, and each
value dist(v ) is the actual shortest-path distance from s to v. In particular, if s
cannot reach v, then dist(v ) = oo, and if any negative cycle is reachable from s,
then the algorithm never terminates.

The correctness of Ford’s generic algorithm follows from the following series
of simpler claims:

r. At any moment during the execution of the algorithm, for every vertex v,
the distance dist(v ) is either oo or the length of a walk from s to v. This
claim can be proved by induction on the number of relaxations.

2. If the graph has no negative cycles, then dist(v ) is either oo or the length
of some simple path from s to v. Specifically, if dist(v ) is the length of
a walk from s to v that contains a directed cycle, that cycle must have
negative length. This claim implies that if G has no negative cycles, the
relaxation algorithm eventually halts, because there are only a finite
number of simple paths in G.

3. If no edge in G is tense, then for every vertex v, the distance dist(v )
is the length of the predecessor path s-> ■ • -pred(pred(v))^pred(v)->v.
Specifically, if v violates this condition but its predecessor pred(v) does
not, the edge pred(v)->v is tense.

4. If no edge in G is tense, then for every vertex v, the path of predecessor

edges s->->pred(pred(v))^pred(v)^v is in fact a shortest path from s

to v. Specifically, if v violates this condition but its predecessor u in some
shortest path does not, the edge u-»v is tense. This claim also implies
that if G has a negative cycle, then some edge is always tense, so the
generic algorithm never halts.

So far I haven’t said anything about how to find tense edges, or which tense
edge to relax if there is more than one. Just as with graph traversal, there are

275

8. Shortest Paths

several different instantiations of Ford’s generic relaxation algorithm. Unli k e
graph traversal, however, the efficiency and correctness of each search strategy
depends on the structure of the input graph.

In the rest of this chapter, we discuss the four most common instantiations
of Ford’s algorithm, each of which is the best choice for a different class of
input graphs. I’ll leave the remaining details of the generic correctness proof as
exercises, and instead give (more informative) correctness proofs for each of
these four specific algorithms.

8.4 Unweighted Graphs: Breadth-First Search

In the simplest special case of the shortest path problem, all edges have weight 1 ,
and the length of a path is just the number of edges. This special case can be
solved by a species of our generic graph-traversal algorithm called breadth-first
search. Breadth-first search is often attributed to Edward Moore, who described
it in 1957 (as “Algorithm A”) as the first published method to find the shortest
path through a maze. 4 Especially in the context of VLSI wiring and robot path
planning, breadth-first search is sometimes attributed to Chin Yang Lee, who
described several applications of Moore’s “Algorithm A” (with proper credit to
Moore) in 1961. However, in 1945, more than a decade before Moore considered
mazes, Konrad Zuse described an implementation of breadth-first search, as a
method to count and label the components of a disconnected graph. 6

4 Moore was motivated by a weakness in Claude Shannon’s maze-solving robot “Theseus”,
which Shannon designed and constructed in 1950. (Theseus used a memoized version of depth-
first search, implemented using electromechanical relays; this was almost certainly the first
implementation of depth-first search in graphs.) According to Moore, “When this machine was
used with a maze which had more than one solution, a visitor asked why it had not been built
to always find the shortest path. Shannon and I each attempted to find economical methods
of doing this by machine. He found several methods suitable for analog computation, 5 and I
obtained these algorithms.”

5 Analog methods for computing shortest paths through mazes have been proposed using
ball bearings, fluid/plasma flow, chemical reaction waves, chemotaxis, resistor networks, electric
circuits with LEDs, memristor networks, glow discharge in microfluidic chips, growing plants,
slime mold, amoebas, ants, bees, nematodes, and tourists.

6 Konrad Zuse was one of the early pioneers of computing; he designed and built his first
programmable computer (later dubbed the Zi) in the late 1930s from metal strips and rods in his
parents’ living room; the Zi and its original blueprints were destroyed by a British air raid in 1944.
Zuse’s 1945 PhD thesis describes the very first high-level programming language, called Plankalkul.
The first complete example of a Plankalkul program in Zuse’s thesis is an implementation of
breadth-first search to count components, along with a pseudocode explanation and an illustrated
step-by-step trace of the algorithm’s execution on a disconnected graph with eight vertices. Due
to the collapse of the Nazi government, Zuse was unable to submit his PhD thesis, and Plankalkill
remained unpublished until 1972. The first Plankalkill compiler was finally implemented in 1975
by Joachim Hohmann.

276

8 . 4 - Unweighted Graphs: Breadth-First Search

Breadth-first search maintains a first-in-first-out queue of vertices, which
initially contains only the source vertex s. At each iteration, the algorithm Pulls
a vertex u from the front of the queue and examines each of its outgoing edges
u-yv. Whenever the algorithm discovers an outgoing tense edge u-»v, it relaxes
that edge and PusHes vertex v onto the queue. The algorithm ends when the
queue becomes empty.

BFS(s):

InitSSSP(s)

Push(s)

while the queue is not empty
u <— Pull( )
for all edges u-w

if dist(v) > dist(u) + 1
dist(v ) <— dist(u) + 1
pred(v) <— u
Push(v)

((ifu^v is tense))
((relax u->v))

Breadth-first search is somewhat easier to analyze if we break its execution
into phases, by introducing an imaginary token. Before we Pull any vertices,
we Push the token into the queue. The current phase ends when we Pull the
token out of the queue; we begin the next phase when we Push the token into
the queue again. Thus, the first phase consists entirely of scanning the source
vertex s. The algorithm ends when the queue contains only the token. The
modified algorithm is shown in Figure 8.5, and Figure 8.6 shows an example of
this algorithm in action. Let me emphasize that these modifications are merely
a convenience for analysis; with or without the token, the algorithm PusHes and
Pulls vertices in the same order, scans edges in the same order, and outputs
exactly the same distances and predecessors.

BFSWithToken(s) :

InitSSSP(s)

Push(s)

Push(*) ((start the first phase))

while the queue contains at least one vertex
u <— Pull( )
if u = *

Push(*) ((start the next phase))

else

for all edges u-w

if dist(v ) > dist(u) + 1
dist(v ) <— dist(u) + 1
pred(v)<— u
Push(v)

((if u^>v is tense))
((relax u->v))

Figure 8.5. Breadth-first search with an end-of-phase token (it); bold red lines are only for analysis.

277

8. Shortest Paths

Figure 8.6. A complete run of breadth-first search in a directed graph. Vertices are pulled from the
queue in the order s^bd^cag^fe^h^^, where * is the end-of-phase token. Bold vertices are
in the queue at the end of each phase. Bold edges describe the evolving shortest path tree.

Let me emphasize that in the following lemma, dist(v ) is just a variable
maintained by the algorithm. While dist(v ) intuitively represents a tentative
shortest-path distance, we cannot assume (yet) that dist(v ) is ever actually equal
to the true shortest-path distance from s to v. Don’t worry; we’ll get there.

Lemma 8.1. For every integer i > 0 and every vertex v, at the end of the i th
phase, either dist{v) = 00 or dist(v) < i, and v is in the queue if and only if
dist(v ) = i.

Proof: The proof proceeds by induction on i. The base case i = 0 is straight¬
forward: At the start of the first phase (“at the end of the zeroth phase”), the
queue contains only the start vertex s and the token *, and InitSSSP just set
dist(s ) <— 0 and dist(v ) <— 00 for all v 7 - s.

So fix an integer i > 0 . The inductive hypothesis implies that at the start of
the ith phase, the queue contains every vertex u with dist(u) = i — 1, followed
by the token *. In other words, the queue looks like this:

-> * i — 1 i — 1 i — 1 ->

Thus, before we Pull the token * from the queue, ending the ith phase, we
Pull every vertex u with dist(u ) = i — 1 .

For each such vertex u, we consider every outgoing edge u->v. If u->v is
tense, we set dist(v) <— dist(u ) + 1, so that dist(v ) = i, and then immediately

278

8 . 4 - Unweighted Graphs: Breadth-First Search

Push v into the queue. These are the only assignments to distance labels
during the ith phase. Thus, by induction, during the entire ith phase, the queue
contains some vertices with distance label i — 1, followed by the token, followed
by some vertices with distance label i:

4 i ••• i * i — 1 i — 1 4

In particular, just before the ith phase ends, the queue contains the token,
followed by some vertices with distance label i.

i i ■ ■ ■ i *

Moreover, vertex v appears in this final queue if and only if dist(v) was changed
during the ith phase. Thus, at the end of the ith phase, the queue contains every
vertex v with dist(v) = i. □

Lemma 8.1 implies that the main body of BFS assigns distance labels in non¬
decreasing order; on the other hand, the distance label dist(v ) of each vertex v
never increases. It follows that for each vertex v, the line “dist(v) <— dist(u) + 1 ”
is executed at most once, during phase dist(v). Similarly:

• Each predecessor pointer pred(v) is changed at most once, during phase
dist(v).

• Each vertex v is PusHed into the queue at most once, during phase dist(v).

• Each vertex u is PuLLed from the queue at most once, during phase dist(u)+ 1 .

• For each edge u->v, the comparison “is dist(v ) > dist(u) + 1 ” is performed
at most once, during phase dist(u ) + 1.

Altogether, these observations imply that breadth-first search runs in 0 (V + E)
time. Intuitively, we can think of the vertices in the queue as a “wavefront”
expanding monotonically outward from the source vertex s, passing over each
vertex and edge of the graph at most once. This expanding wavefront analogy
was already proposed by Chin Yang Lee in 1961, inspired by visualizations
produced by his implementation of Moore’s Algorithm A.

These observations also imply that we can replace the condition “if dist(v) >
dist(u ) + 1 ” by the (arguably) simpler test “if dist(v) = 00”. Then distances
play the same role as the marks maintained by other graph-traversal algorithms,
which ensure that each vertex is visited only once. Specifically, a vertex is
“marked” if and only if its distance label is finite.

But we still need to prove that the final distance labels are correct!

Theorem 8 . 2 . When BFS ends, dist( v) is the length of the shortest path in G
from s to v, for every vertex v.

279

8. Shortest Paths

Proof: Fix an arbitrary vertex v, and consider an arbitrary path Vg-nq-*- >v f

in G, where v 0 = s and V( = v. I claim that dist(vj) < j for each index j; in
particular dist(v) < l. We can prove this claim by induction on j as follows.

• Trivially dist(v 0 ) = dist(s) = 0 .

• For any index j > 0 , the induction hypothesis implies dist^Vj^) < j — 1 .
Immediately after we Pull vertex v J -_ 1 from the queue, either dist^v^ <
dist(Vj_i ) + 1 already, or we set dist(v ; ) <— dist(vj_i ) + 1 . In either case, we
have dist^Vj ) < dist(vj_i ) + 1 < j.

We just proved that dist(v ) is at most the length of an arbitrary path from s to v;
it follows that dist(v ) is at most the length of the shortest path from s to‘v.

A similar induction proof implies that dist(v) is the length of the predecessor
path s-y - >pred(pred(v))^pred(v)->v , so this must be the shortest path. □

8.5 Directed Acyclic Graphs: Depth-First Search

Shortest paths are also easy to compute in directed acyclic graphs, even when
the edges are weighted, and in particular, even when some edges have negative
weight. (We don’t have to worry about negative cycles, because by definition,
dags don’t have any cycles!) Indeed, this is a completely standard dynamic
programming algorithm.

Let G be a directed graph with weighted edges, and let s be the fixed start
vertex. For any vertex v, let dist(v) denote the length of the shortest path in G
from s to v. This function satisfies the following simple recurrence:

In fact, this identity holds for all directed graphs, but it is only a recurrence
for directed acyclic graphs. If the input graph G contained a cycle, a recursive
evaluation of this function would fall into an infinite loop; however, because G
is a dag, each recursive call visits an earlier vertex in topological order.

The dependency graph for this recurrence is the reversal of the input graph G:
subproblem dist(v) depends on dist(v ) if and only if u->v is an edge in G. Thus,
we compute the distance of every in 0 (V + E) time by performing a depth-first
search in the reversal of G and considering vertices in postorder. Equivalently,
we can consider the vertices in the original graph G in topological order, as
shown in Figure 8.7.

The resulting dynamic-programming algorithm is another example of Ford’s
generic relaxation algorithm! To make this connection clearer, we can move the
initialization dist(v ) outside the main loop and add computation of predecessor
pointers, as shown in Figure 8.8. Figure 8.10 shows this algorithm in action.

280

8.5. Directed Acyclic Graphs: Depth-First Search

DagSSSP(s):

for all vertices v in topological order

if v = s

dist(v ) <— 0

else

dist(v) <— 00

for all edges u-rv

if dist(v ) > dist(u) + w(u-^v)

(( ifu->v is tense))

dist(v ) <— dist(u) + w(u->v)

((relax u->v))

Figure 8.7. Computing shortest paths in a dag using dynamic programming

DagSSSP(s):

InitSSSP(s)

for all vertices v in topological order
for all edges u^>v
if u-rv is tense
Relax(u->v)

Figure 8.8. Computing shortest paths in a dag using Ford's algorithm. (These are the same algorithm.)

Figure 8.9. Computing shortest paths in a dag, by relaxing incoming edges in topological order. In
each iteration, bold edges indicate predecessors, and the bold vertex is about to be scanned. Compare
with Figure 8.10.

281

8. Shortest Paths

DagSSSP differs from breadth-first search and other instances of Ford’s
relaxation strategy in one minor respect. Whenever these other shortest-path
algorithms consider a vertex, they attempt to relax each of its outgoing edges,
intuitively pushing the wavefront outward; whereas, DagSSSP attempts to relax
each of the incoming edges of each vertex, intuitively pulling the wavefront
forward.

However, if we modify DagSSSP to relax outgoing edges instead of incoming
edges, we obtain another algorithm that computes shortest paths in dags in
0 (V + E) time and that more closely resembles other shortest-path algorithms.

PushDagSSSP(s):

InitSSSP(s)

for all vertices u in topological order
for all outgoing edges u-w
if u—>v is tense
Relax(u^v)

Figure 8.10 shows an execution of this modified algorithm on the same
graph as Figure 8.9. The correctness of PushDagSSSP follows immediately
from the correctness of Ford’s general relaxation strategy, but it’s not hard to
prove correctness directly, by induction over the vertices in topological order.

282

Figure 8.10. Computing shortest paths in a dag, by relaxing outgoing edges in topological order. In
each iteration, bold edges indicate predecessors, and the bold vertex is about to be scanned. Compare
with Figure 8.g.

8.6. Best-First: Dijkstra's Algorithm

8.6 Best-First: Dijkstra's Algorithm

If we replace the FIFO queue in breadth-first search with a priority queue, where
the key of a vertex v is its tentative distance dist(v), we obtain an algorithm first
“published” in 1957 by a team of researchers at the Case Institute of Technology,
in an annual project report for the Combat Development Department of the
US Army Electronic Proving Ground. The same algorithm was independently
discovered by Edsger Dijkstra in 1956 (but not published until 1959), again by
George Minty some time before i960, and again by P. D. Whiting and J. A. Hillier
in i960. A nearly identical algorithm was also described by George Dantzig in
1958. Although several early sources called it “Minty’s algorithm”, this approach
is now universally known as “Dijkstra’s algorithm”, in full accordance with
Stigler’s Law. 7 Pseudocode for this algorithm is shown in Figure 8.11.

Dijkstra(s):

InitSSSP(s)

Insert(s, 0)

while the priority queue is not empty
u <— ExtractMin( )
for all edges u-w
if u—>v is tense

Relax(u^v)

if v is in the priority queue
DecreaseKey(v, dist(v))

else

Insert(v, dist(v ))

Figure 8.11. Dijkstra's algorithm.

An easy induction proof implies that, at all times during the execution of this
algorithm, an edge u->v is tense if and only if vertex u is either in the priority
queue or is the vertex most recently ExTRACTed from the priority queue. Thus,
Dijkstra’s algorithm is an instance of Ford’s general strategy, which implies that
it correctly computes shortest paths, provided there are no negative cycles in G.

No Negative Edges

Dijkstra’s algorithm is particularly well-behaved when the input graph has no
negative-weight edges. In this setting, the algorithm intuitively expands a
wavefront outward from the source vertex s, passing over vertices in increasing

7 I will follow this common convention, despite the historical inaccuracy, partly because I
don’t think anybody wants to read about the “Leyzorek-Gray-Johnson-Ladew-Meaker-Petry-Seitz-
Dantzig-Dijkstra-Minty-Whiting-Hillier algorithm”, and partly because papers that aren’t actually
published don’t count.

283

8. Shortest Paths

order of their distance from s, similarly to breadth-first search. Figure 8.12
shows the algorithm in action.

Figure 8.12. The first four iterations of Dijkstra’s algorithm on a graph with no negative edges. In each
iteration, bold edges indicate predecessors; shaded vertices are in the priority queue; and the bold vertex
is about to be scanned. The remaining iterations do not change the distances or the shortest-path tree.

We can derive a self-contained proof of correctness for Dijkstra’s algorithm
in this setting by formalizing this wavefront intuition. For each integer i, let u ;
denote the vertex returned by the ith call to ExtractMin, and let d ; be the
value of dist(iq) just after this ExTRACTion. In particular, we have u 1 = s and
dj = 0. We cannot assume at this point that the vertices iq are distinct; in
principle, the same vertex might be ExTRACTed more than once.

Lemma 8.3. If G has no negative-weight edges, then for all i < j, we have
di < dj.

Proof: Assume G has no negative weight edges. Fix an arbitrary index i; to
prove the lemma, it suffices to prove that d i+1 > dj. There are two cases to
consider.

• Suppose the edge tq-nq+i is relaxed during the ith iteration of the main
loop. Then at the end of the ith iteration, we have dist(u i+1 ) = dist(iq) +
w(iq-nq +1 ) > dist( iq), because all edge weights are non-negative.

• On the other hand, suppose the edge iq-nq +1 is not relaxed during the
ith iteration of the main loop. Then at the start of the ith iteration, u i+1
must already be in the priority queue, with priority dist(u i+1 ) > dist (iq),
because iq is the vertex returned by ExtractMin. Moreover, dist(u i+1 ) does
not change during the ith iteration.

284

8 .6. Best-First: Dijkstra's Algorithm

In both cases, we conclude that d i+1 > d r The lemma now follows immediately
by induction on i. □

Lemma 8.4. If G has no negative-weight edges, each vertex of G is ExTRACTed
from the priority queue at most once.

Proof: Suppose v is ExTRACTed more than once. Specifically, suppose v is
ExTRACTed in the ith iteration of the main loop, relNSERTed during the j th
iteration, and reExTRACTed during the /cth iteration, for some indices i < j <k.
Then in the notation of the previous proof, we have v = u t = u k .

The distance label dist(v ) never increases. Moreover, dist(v ) strictly decreases
during the jth iteration, just before v is relNSERTed. It follows that d L > d k .
Therefore, by the previous lemma, G has at least one negative-weight edge. □

Lemma 8.4 immediately implies that each vertex is scanned at most once,
and thus that each edge is relaxed at most once. However, unlike in breadth-first
search, each distance label dist(v ) can change multiple times. The first time
dist(v ) changes from 00, we Insert v into the priority queue; after that, each
change to dist(v ) is followed by a call to DecreaseKey. After v is ExTRACTed
from the priority queue, its distance label never changes.

The rest of the correctness proof is almost identical to breadth-first search.

Theorem 8.5. If G has no negative-weight edges, then when Dijkstra ends,
dist(v) is the length of the shortest path in G from s to v, for every vertex v.

Proof: Fix an arbitrary vertex v, and consider an arbitrary path -

in G, where v 0 = s and v k = v. For any index j, let L , denote the length of the
subpath Vo^Vj-*- >Vj. We prove by induction that dist(Vj) < Lj for all j.

• Trivially dist(v 0 ) = dist(s) = 0 = L 0 .

• For any index j > 0 , the induction hypothesis implies dist^Vj^) < L } _,.
Immediately after we Pull vertex v J _ 1 from the queue, either di.st(v ; ) <
dist(vj_ 1 ) + w(vj_ 1 ^v ; ) already, or we set dist^v^ <— dist(v ; _ 1 ) + w(v ; _ 1 ^v ; ).
In either case, we have

dist(vj ) < dist(vj_!)+ w(vj_i^Vj) < Lj_i+w(vj_i~>Vj) = Lj.

We just proved that dist(v ) is at most the length of an arbitrary path from s to v;
it follows that dist(v) is at most the length of the shortest path from s to‘v.

On the other hand, a similar induction proof implies that dist(v ) is the length
of the predecessor path - >pred(pred(v))^pred(v)->v. □

285

8. Shortest Paths

It remains only to bound the algorithm’s running time. Altogether Dijkstra
performs at most E DecreaseKey operations, and at most V Insert and
ExtractMin operations. Thus, if we implement the underlying priority queue
using a standard binary heap, which supports each operation in O(logV) time,
Dijkstra runs in O(ElogV) time . 8

If we know in advance that our input graphs will never have negative edges,
we can simplify Dijkstra’s algorithm slightly, by iNSERTing every vertex into the
priority queue in the initialization phase, and then only calling DecreaseKey in
the main loop, as shown in Figure 8.13. This is the version of Dijkstra’s algorithm
presented by most algorithms textbooks, Wikipedia, and even Dijkstra’s original
paper; it’s also the version of Dijkstra’s algorithm that I described as “best-first
search” in Chapter 5.

NqnnegativeDi jkstra(s ):

InitSSSP(s)
for all vertices v

Insert(v, dist(v ))
while the priority queue is not empty
u <— ExtractMin( )
for all edges u^v
if u-tv is tense
Relax(u^v)

DecreaseKey(v, dist(_v ))

Figure 8.13. Dijkstra's algorithm very slightly simplified for graphs without negative edges. Differences
from Dijkstra are bold red.

^Negative Edges

However, NonnegativeDijkstra does not correctly compute shortest paths
in graphs with negative edges. Moreover, even when all edge weights are
positive, NonnegativeDijkstra is no faster than Dijkstra (either in theory
or in practice). For both of these reasons, I think Dijkstra is more deserving
of the name “Dijkstra’s algorithm” than NonnegativeDijkstra. Even Edsger
Dijkstra would have agreed that a correct algorithm that is sometimes (and in
practice, rarely) slow is better than a fast algorithm that doesn’t always work!

8 Shortest-path papers from the 1950s never mentioned priority queues. Dijkstra proposed a
brute-force scan of all vertices on the wavefront at every iteration; his original algorithm runs
in 0 (V 2 ) time, which is actually faster than the binary-heap implementation when E = D(V 2 )!
Minty proposed a brute-force scan of all edges u->v such that dist(u ) is finite but dist(v ) is not;
thus, his original algorithm runs in 0 (VE ) time. The use of a priority queue, implemented as
a binary heap, to obtain near-linear running time was proposed by Donald Johnson in 1977.
The running time can be improved to 0 (£ + VlogV) using a more complex priority queue data
structure called a Fibonacci heap. There are even faster algorithms, using even more sophisticated
priority queues, for the special case of integer edge weights.

286

8 . 7 - Relax ALL the Edges: Bellman-Ford

Unfortunately, when the input graph has negative edges, the familiar
“expanding wavefront” intuition is no longer accurate. The same vertex could
be ExTRACTed multiple times; the same edge could be relaxed multiple times;
and distances might not be discovered in increasing order. Figure 8.15 shows an
example execution where the top left vertex is Extracted six times, and the
top three edges are each relaxed twice.

For graphs without negative cycles, but no other restrictions on edge weights,
the worst-case running time of Dijkstra is actually exponential. Figure 8.14
shows particularly simple family of graphs (due to Douglas Shier and Christoph
Witzgall) that forces Dijkstra to perform 0 ( 2 v ^ 2 ) relaxations. 9 A more complex
family of graphs (which I’ll leave as an exercise) forces 0 ( 2 l ) relaxations, which
is the worst possible. In practice, however Dijkstra’s algorithm is usually quite
fast even for graphs with negative edges.

Figure 8.14. A directed graph with negative edges that forces Dijkstra to run in exponential time.

8.7 Relax ALL the Edges: Bellman-Ford

The simplest implementation of Ford’s generic shortest-path algorithm was
first sketched by Alfonso Shimbel in 1954, described in more detail by Edward
Moore in 1957, and independently rediscovered by Max Woodbury and George
Dantzig in 1957, by Richard Bellman in 1958, and by George Minty in 1958.
(Neither Woodbury and Dantzig nor Minty published their algorithms.) In full
compliance with Stigler’s Law, the algorithm is almost universally known as
Bellman-Ford , 10 because Bellman explicitly used Ford’s 1956 formulation of
relaxing edges, although some authors refer to “Bellman-Kalaba” 11 and a few
early sources refer to “Bellman-Shimbel”.

9 Amusingly, Shier and Witzgall’s example is a dag with only O(V) edges, which means
shortest paths can be computed in 0 (V ) time, even if we didn’t already notice that the zig-zag
path along the top is the shortest path tree.

“I will follow this common convention, despite the historical inaccuracy, partly because I
don’t think anyone really wants read about the “Shimbel/Moore/Woodbury-Dantzig/Bellman-
Ford/Kalaba/Minty algorithm”, and partly because I’m tired of people looking at me funny
when I talk about “Shimbel’s algorithm”.

“This name is most likely a reference to Richard Bellman and Robert Kalaba’s 1965 monograph
on dynamic programming and control theory, which describes Bellman’s algorithm. Bellman and
Kalaba also published an extension of Bellman’s algorithm in i960 that computes fcth shortest
paths, for any constant k.

287

8. Shortest Paths

Figure 8.15. A complete run of Dijkstra's algorithm on a graph with negative edges. At each iteration,
bold edges indicate predecessors; shaded vertices are in the priority gueue; and the bold vertex is the
next to be scanned. Compare with Figure 8.17.

288

8 . 7 - Relax ALL the Edges: Bellman-Ford

The Shimbel / Moore / Woodbury-Dantzig / Bellman-Ford / Kalaba / Minty /
Brosh 12 algorithm can be summarized in one line:

Bellman-Ford: Relax ALL the tense edges, then recurse.

BellmanFqrd(s)

InitSSSP(s)

while there is at least one tense edge
for every edge u-w
if it—>v is tense
Relax(u-w)

The following lemma is the key to proving both correctness and efficiency
of Bellman-Ford. For every vertex v and non-negative integer i, let dist<i(v )
denote the length of the shortest walk in G from s to v consisting of at most i
edges. In particular, dist< 0 (s) = 0 and dist< 0 (v ) = oo for all v 7 -s.

Lemma 8.6. For every vertex v and non-negative integer i, after i iterations of
the main loop of BellmanFord, we have dist(v) < dist^v).

Proof: The proof proceeds by induction on i. The base case i = 0 is trivial,
so assume i > 0 . Fix a vertex v, and let W be the shortest walk from s to v
consisting of at most i edges (breaking ties arbitrarily). By definition, W has
length dist^v). There are two cases to consider.

• Suppose W has no edges. Then W must be the trivial walk from s to s, so
v = s and dist^s) = 0 . We set dist(s ) <— 0 in InitSSSP, and dist(s ) can
never increase, so we always have dist(s ) < 0.

• Otherwise, let u->v be the last edge of W. The induction hypothesis implies
that after i — 1 iterations, dist(u ) < dist^^u). During the ith iteration of
the outer loop, when we consider the edge u-»v in the inner loop, either
dist(v ) < dist(u ) + w(u->-v) already, or we set dist(v ) <— dist(u ) + w(u->-v).
In both cases, we have dist(v ) < dist< i _±(u) + w(u^>-v') = dist^v). As usual,
dist(v ) cannot increase (although dist(v ) might decrease further before the
ith iteration of the outer loop ends).

In both cases, we conclude that dist(v ) < dist^v) at the end of the ith
iteration. □

If the input graph has no negative cycles, the shortest walk from s to any
other vertex is a simple path with at most V — 1 edges; it follows that Bellman-
Ford halts with the correct shortest-path distances after at most V — l iterations.

12 Go read everything in Hyperbole and a Half again. And then adopt another cat, so you can
buy it another copy of the book.

289

8. Shortest Paths

Said differently, if any edge is still tense after V — 1 iterations, then the input
graph must contain a negative cycle! Thus, we can rewrite the algorithm more
concretely as follows:

BellmanFord(s)

InitSSSP(s)
repeat V — 1 times

for every edge u^v
if u—>v is tense
Relax(u^v)
for every edge u^v
if u->v is tense

return “Negative cycle!”

Each iteration of the inner loop trivially requires 0 (E) time, so the overall
algorithm runs in O(VE) time. Thus, Bellman-Ford is always efficient, even if
the graph has negative edges, and in fact even if the graph has negative cycles.

If all edge weights are non-negative, however, Dijkstra’s algorithm is faster,
at least in the worst case. (In practice, Dijkstra’s algorithm is often faster than
Bellman-Ford even for graphs with negative edges.)

Moore’s Improvement

Neither Moore nor Bellman described the Bellman-Ford algorithm in the form I’ve
presented here. Moore presented his version of the algorithm ("Algorithm D") in
the same paper that proposed breadth-first search ("Algorithm A') for unweighted
graphs; indeed, the two algorithms are nearly identical. Although Moore’s
algorithm has the same O(VE) worst-case running time as BellmanFord, it is
often significantly faster in practice, intuitively because it avoids checking edges
that are “obviously” not tense.

Moore derived his weighted shortest-path algorithm by making two modifi¬
cations to breadth-first search. First, replace each “+1” with “+w(iz^v)" in the
innermost loop, to take the edge weights into account. Second, check whether
a vertex is already in the FIFO queue before iNSERTing it, so that the queue
always contains at most one copy of each vertex. 13

Following our earlier analysis of breadth-first search, I’ll introduce a “token”
to break the execution of the algorithm into phases. Just like BFS, each phase
begins when the token is PusHed into the queue, and ends when the token
is PuLLed out of the queue again. Just like BFS, the algorithm ends when the
queue contains only the token. The resulting algorithm is shown in Figure 8.16.

Because the queue contains at most one copy of each vertex at any time,
each vertex is PuLLed from the queue at most once in each phase, and therefore

13 Moore’s algorithm is still correct without this check, but the 0 (VE ) time bound is not.

290

8 . 7 - Relax ALL the Edges: Bellman-Ford

Mqore(s):

InitSSSP(s)

Push(s)

Push(*) ((start the first phase))

while the queue contains at least one vertex
u <— Pull( )
if u = *

Push(*) ((start the next phase))

else

for all edges u^v
if u—is tense
Relax(u^v)

if v is not already in the queue
Push(v)

Figure 8.16. Moore's shortest-path algorithm. Bold red lines involving the token * are only for analysis.

each edge u->v is checked for tenseness at most once in each phase. Moreover,
every edge that is tense when a phase begins is relaxed during that phase.
(Some edges that become tense during the phase might also be relaxed during
that phase, and some relaxed edges might become tense again in the same
phase.) Thus, Moore can be viewed as a refinement of BellmanFord that
uses a queue to maintain tense edges, rather than testing every edge by brute
force. In particular, a similar inductive proof establishes the following analogue
of Lemma 8.6:

Lemma 8.7. For every vertex v and non-negative integer i, after i phases of
Moore, we have dist(v ) < dist^v).

Thus, if the input graph has no negative cycles, Moore halts after at most
V — 1 phases. In each phase, we scan each vertex at most once, so we relax
each edge at most once, so the worst-case running time of a single phase is
0 (E). Thus, the overall running time of Moore is 0 (VE). In practice, however,
Moore often computes shortest paths considerably faster than BellmanFord,
because it only scans an edge u->v if dist(u) was changed in the previous phase.

If the input graph contains a negative cycle, Moore never halts. Fortunately,
like BellmanFord, it is easy to modify Moore’s algorithm to report negative
cycles if they exist. Perhaps the easiest modification is to actually maintain a
token, and count the number of times the token is PuLLed from the queue. Then
the input graph contains a negative cycle if and only if the queue is non-empty
immediately after the token is PuLLed for the (V — l)th time.

8. Shortest Paths

*

Figure 8.17. A complete run of Moore's algorithm on a directed graph with negative edges. Nodes are
pulled from the queue in the orders *aba<df>i<dc'i<d**, where * is the end-of-phase token. At
the start of each phase, bold edges indicate predecessors, and shaded vertices are in the vertex queue.
Compare with Figures 8.6 and 8.15.

Dynamic Programming Formulation

Like almost everything else with his name on it, Richard Bellman derived the
“Bellman-Ford” shortest-path algorithm via dynamic programming. As usual, we
need to start with a recursive definition of shortest path distances. It’s tempting
to use the same identity that we exploited for directed acyclic graphs:

Jo if v = s

dist(v) | m j n -1- w(u->v)) otherwise

Unfortunately, if the input graph is not a dag, this recurrence doesn’t work!

Suppose the input graph contains the directed cycle u->v->w->u. To compute
dist(w ) we first need dist(v), and to compute dist(v ) we first need dist(u), but
to compute dist(u ) we first need dist(w). If the input graph has any directed
cycles, we get stuck in an infinite loop!

To support a proper recurrence, we need to add an additional structural
parameter to the distance function, which decreases monotonically at each
recursive call, defined so that the function is trivial to evaluate when the

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8 . 7 - Relax ALL the Edges: Bellman-Ford

parameter reaches 0 . Bellman chose the maximum number of edges as this
additional parameter. 14

As in our earlier analysis, let dist<fv) denote the length of the shortest walk
from s to v consisting of at most i edges. Bellman observed that this function
obeys the following Bellman’s equation recurrence:

r

0

di5t<i(v)

oo

min

dist<i_i(v)

min (dist^i-^u) + w(u->v ))

if t = 0 and v = s
if i = 0 and v/s

otherwise

Let’s assume that the graph has no negative cycles, so our goal is to compute
dist^ v _ 1 (v) for every vertex v. Here is a straightforward dynamic-programming
evaluation of this recurrence, where dist[i, v] stores the value of dist<fv).
Correctness of the final shortest-path distances follows from the correctness of
the recurrence, and the 0 (VE ) running time is obvious. This is essentially how
Bellman presented his shortest-path algorithm.

BellmanFqrdDP(s)
dist[0,s] <— 0
for every vertex v f s
dist[ 0, v ] <— oo
for i <— 1 to V — 1
for every vertex v

dist[i, v] <— dist[i — 1, v]
for every edge u->v

if dist[i, v] > dist[i — 1, u] + w(u->v)
dist[i, v] <— dist[i — l,u] + w(u-tv)

We can transform this dynamic programming algorithm into our original
formulation of BellmanFord through a short series of minor modifications.
First, each iteration of the outermost loop considers each edge u^v exactly once,
but the order in which we consider those edges doesn’t actually matter. Thus,
we can safely remove one level of indentation from the last three lines! The
modified algorithm may consider edges in a different order, but it still correctly
computes dist<fv) for all i and v.

14 As we’ll see in the next chapter, this is not the only reasonable choice.

293

8. Shortest Paths

BellmanFordDP2(s)
dist[0,s] <— 0
for every vertex v ^ s
dist[ 0, v] <— oo
for i «— 1 to V — 1
for every vertex v

dist[i, v] <— dist[i — 1, v]
for every edge u->v

if dist[i, v] > dist[i — 1, u] + w(u^v)
dist[i, v] <— dist[i — 1,u] + w(u->v)

Next we change the indices in the last two lines from i — 1 to i. This
change may cause the distances dist[i,v] to approach the true shortest-path
distances more quickly than before, but the algorithm correctly computes the
true shortest path distances. Instead of dist[i,v ] = dist< i (v'), we now have
dist[i, v] < dist<j(v) for all i and v, mirroring Lemmas 8.6 and 8.7.

BellmanFordDP3(s)
dist[0,s] <— 0
for every vertex v/s
dist[ 0, v] <— 00
for i <— 1 to V — 1
for every vertex v

dist[i, v] <— dist[i — 1, v]
for every edge u->v

if dist[i, v] > dist[i,u ] + w(u^v) ((not i — 1/))
dist[i, v] <— dist[i,u ] + w(u->v) ((not i — 1/))

But this algorithm is a little silly. In the ith iteration of the outermost loop, we
first copy the (i — l)th row of the array dist [*, *] to the ith row, and then modify
the elements of the ith row. So we really don’t need a two-dimensional array at
all; the iteration index i is completely redundant! In our final modification, we
maintain only a one-dimensional array of tentative distances.

BellmanFqrdFinal(s)
dist[s] <— 0
for every vertex v ^ s
dist[v] <— 00
for i <— 1 to V — 1

for every edge u^>v

if dist[v] > dist[u] + w(u-vv)
dist[v] <— dist[u] + w(u^v)

294

This final dynamic programming algorithm is almost identical to our original
formulation of BellmanFord! The first three lines initialize the shortest path
distances, and the last two lines relax the edge u->v if that edge is tense.

Exercises

BellmanFordFinal is missing only two features of our earlier formulation: It
does not maintain predecessor pointers or detect negative cycles. Fortunately,
adding those features is straightforward.

Exercises

o. Let G be a directed graph with arbitrary edge weights (which maybe positive,
negative, or zero), possibly with negative cycles, and let s be an arbitrary
vertex of G.

(a) Suppose every vertex v stores a number dist(v ) (but no predecessor
pointers). Describe and analyze an algorithm to determine whether
dist(v) is the shortest-path distance from s to v, for every vertex v.

(b) Suppose instead that every vertex v 7 - s stores a pointer pred(v) to
another vertex in G (but no distances). Describe and analyze an
algorithm to determine whether these predecessor pointers define a
single-source shortest path tree rooted at s.

1. A looped tree is a weighted, directed graph built from a binary tree by adding
an edge from every leaf back to the root. Every edge has non-negative
weight.

A looped tree.

(a) How much time would Dijkstra’s algorithm require to compute the
shortest path between two vertices u and v in a looped tree with n
nodes?

(b) Describe and analyze a faster algorithm.

2. Suppose we are given a directed graph G with weighted edges and two
vertices s and t.

(a) Describe and analyze an algorithm to find the shortest path from s to
t when exactly one edge in G has negative weight. [Hint: Modify
Dijkstra’s algorithm. Or don’t.]

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8. Shortest Paths

(b) Describe and analyze an algorithm to find the shortest path from s to t
when exactly k edges in G have negative weight. How does the running
time of your algorithm depend on k ?

3. Suppose we are given an undirected graph G in which every vertex has a
positive weight.

(a) Describe and analyze an algorithm to find a spanning tree of G with
minimum total weight. (The total weight of a spanning tree is the sum
of the weights of its vertices.)

(b) Describe and analyze an algorithm to find a path in G from one given
vertex s to another given vertex t with minimum total weight. (The total
weight of a path is the sum of the weights of its vertices.)

[Hint: One of these problems is trivial.]

4. For any edge e in any graph G, let G\e denote the graph obtained by
deleting e from G. Suppose we are given a graph G and two vertices s
and t. The replacement paths problem asks us to compute the shortest-path
distance from s to t in G \ e, for every edge e of G. The output is an array
of E distances, one for each edge of G.

(a) Suppose G is a directed graph, and the shortest path from vertex s to
vertex t passes through every vertex of G. Describe an algorithm to solve
this special case of the replacement paths problem in O(ElogV) time.

T (b) Describe an algorithm to solve the replacement paths problem for
arbitrary undirected graphs in 0 {E log V) time.

In both subproblems, you may assume that all edge weights are non-negative.
[Hint: If we delete an edge of the original shortest path, how do the old
and new shortest paths overlap?]

5. Let G = (V, E ) be a connected directed graph with non-negative edge weights,
let s and t be vertices of G, and let H be a subgraph of G obtained by deleting
some edges. Suppose we want to reinsert exactly one edge from G back
into H, so that the shortest path from s to t in the resulting graph is as short
as possible. Describe and analyze an algorithm that chooses the best edge
to reinsert, in O(ElogV) time.

6. (a) Describe and analyze a modification of Bellman-Ford that actually returns

a negative cycle if any such cycle is reachable from s, or a shortest-path
tree if there is no such cycle. The modified algorithm should still run in
0 (VE ) time.

296

Exercises

(b) Describe and analyze a modification of Bellman-Ford that computes the
correct shortest path distances from s to every other vertex of the input
graph, even if the graph contains negative cycles. Specifically, if any
walk from s to v contains a negative cycle, your algorithm should end
with dist(v ) = —co; otherwise, dist(v ) should contain the length of the
shortest path from s to v. The modified algorithm should still run in
0 {VE ) time.

r (c) Repeat parts (a) and (b), but for Ford’s generic relaxation algorithm.
You may assume that the unmodified algorithm halts in 0 ( 2 y ) steps if
there is no negative cycle; your modified algorithms should also run in
0 (2 V ) time.

7. Consider the following even looser variant of Ford’s generic relaxation
algorithm:

FellmanBored(s):

InitSSSP(s)

repeat

e, <— any edge in G
if e, is tense
RELAXfe,)

until you get bored or whatever

Prove that if FellmanBored examines the edges of any walk W starting
from s, in order along W, then the last distance label in W is at most

the length of W. More formally: If the edges of any walk - >v ( ,

where v 0 = s, define a subsequence of the edges e 1 ,e 2 ,e 3 ,... examined
by FellmanBored, then we have distal) < 2 ;=i w ( v ;-i^ v ;)- [Hint: This
property is almost easier to prove than it is to state correctly.]

8. This problem considers several ways to detect negative cycles using Ford’s
generic relaxation algorithm.

(a) Prove that if pred(s) ever changes after InitSSSP, then the input graph
contains a negative cycle through s.

(b) Show that pred(s ) might never change after InitSSSP, even when the
input graph contains a negative cycle through s.

(c) Let P denote the current graph of predecessor edges pred(v)^v, and
let X denote the set of all currently tense edges; both of these sets evolve
as the algorithm executes. Prove that the input graph has no negative
cycles if and only if P U X is always a dag.

(d) Let R denote the set of all edges that have been relaxed so far; this set
grows as the algorithm executes. Prove that the input graph has no
negative cycles if and only if R is always a dag.

297

8. Shortest Paths

r 9. Prove that Dijkstra’s algorithm performs Q( 2 V ) relaxations in the worst case
when edges are allowed to have negative weight, even if the underlying
graph is acyclic. Specifically, for every positive integer n, construct a n-vertex
dag G n with weighted edges, such that Dijkstra’s algorithm calls Relax
ff( 2 ") times when G n is the input graph. [Hint: Towers of Hanoi.]

v io. Prove that Ford’s generic relaxation algorithm (and therefore Dijkstra’s
algorithm) halts after at most 0(2 V ) relaxations, unless the input graph
contains a negative cycle. [Hint: See Problem 8 (d).]

n. Although we typically speak of “the” shortest path between two nodes,
single graph could contain several minimum-length paths with the same
endpoints. Even for weighted graphs, it is often convenient to choose a
minimum-weight path with the fewest edges; call this a best path from s
to t. Suppose we are given a directed graph G with positive edge weights
and a source vertex s in G. Describe and analyze an algorithm to compute
best paths in G from s to every other vertex.

Figure 8.18. Four (of many) equal-length shortest paths. The first path is the "best” shortest path.

12. Describe and analyze an algorithm to determine the number of shortest
paths from a source vertex s to a target vertex t in an arbitrary directed
graph G with weighted edges. You may assume that all edge weights are
positive and that all necessary arithmetic operations can be performed in
0 ( 1 ) time. [Hint: Compute shortest path distances from s to every other
vertex. Throw away all edges that cannot be part of a shortest path from s
to another vertex. What’s left?]

13. You just discovered your best friend from elementary school on Twitbook.
You both want to meet as soon as possible, but you live in two different
cites that are far apart. To minimize travel time, you agree to meet at an
intermediate city, and then you simultaneously hop in your cars and start
driving toward each other. But where exactly should you meet?

You are given a weighted graph G = (V, E), where the vertices V represent
cities and the edges E represent roads that directly connect cities. Each
edge e has a weight w(e) equal to the time required to travel between the

298

Exercises

two cities. You are also given a vertex p, representing your starting location,
and a vertex q, representing your friend’s starting location.

Describe and analyze an algorithm to find the target vertex t that allows
you and your friend to meet as quickly as possible.

14. You are hired as a cyclist for the Giggle Highway View project, which will
provide street-level images along the entire US national highway system.
As a pilot project, you are asked to ride the Giggle Highway-View Fixed-
Gear Carbon-Fiber Bicycle from “the Giggleplex” in Portland, Oregon to
“Gigglesburg” in Williamsburg, Brooklyn, New York.

You are a hopeless caffeine addict, but like most Giggle employees
you are also a coffee snob; you only drink independently roasted organic
shade-grown single-origin espresso. After each espresso shot, you can bike
up to L miles before suffering a caffeine-withdrawal migraine.

Giggle helpfully provides you with a map of the United States, in the
form of an undirected graph G, whose vertices represent coffee shops that
sell independently roasted organic shade-grown single-origin espresso, and
whose edges represent highway connections between them. Each edge e
is labeled with the length ((e) of the corresponding stretch of highway.
Naturally, there are espresso stands at both Giggle offices, represented by
two specific vertices s and t in the graph G.

(a) Describe and analyze an algorithm to determine whether it is possible
to bike from the Giggleplex to Gigglesburg without suffering a caffeine-
withdrawal migraine.

(b) You discover that by wearing a more expensive fedora, you can increase
the distance L that you can bike between espresso shots. Describe
and analyze and algorithm to find the minimum value of L that allows
you to bike from the Giggleplex to Gigglesburg without suffering a
caffeine-withdrawal migraine.

(c) When you report to your supervisor (whom Giggle recently hired away
from their competitor Unter) that the ride is impossible, she demands
to look at your map. “Oh, I see the problem; there are no Starbucks on
this map!” As you look on in horror, she hands you an updated graph G'
that includes a vertex for every Starbucks location in the United States,
helpfully marked in Starbucks Green (Pantone® 3425 C).

Describe and analyze an algorithm to find the minimum number
of Starbucks locations you must visit to bike from the Giggleplex to
Gigglesburg without suffering a caffeine-withdrawal migraine. More
formally, your algorithm should find the minimum number of green
vertices on any path in G' from s to t that uses only edges of length at
most L.

299

8. Shortest Paths

15. Suppose you are given a directed graph G = (V,E) with non-negatively
weighted edges and two vertices s and t. Describe and analyze an algorithm
to find the shortest walk in G from s to t (possibly repeating vertices and/or
edges) whose number of edges is divisible by 3 .

For example, given the graph shown below, with the indicated vertices s
and t, and with all edges having weight 1, your algorithm should return 6,
which is the length of the walks^w^y-s>x-s>s^w->t has length 6.

16. Suppose you are given a directed graph G with non-negatively weighted
edges, where some edges are red and the remaining edges are blue. Describe
an algorithm to find the shortest walk in G from one vertex s to another
vertex t in which no three consecutive edges have the same color. That is, if
the walk contains two red edges in a row, the next edge must be blue, and if
the walk contains two blue edges in a row, the next edge must be red.

For example, given the following graph as input, where every red
edge has weight 1 and every blue edge has weight 2, your algorithm
should return the integer 9 , because the shortest legal walk from s to t is
s^>a^>b^>d—>c=>a^>b—>c.

17. Consider a directed graph G, where each edge has a non-negative weight,
and each edge is colored either red, white, or blue. A walk in G is called
a French flag walk if its sequence of edge colors is red, white, blue, red,

white, blue, and so on. More formally, a walk v 0 ^v 1 -s-- >v k is a French

flag walk if, for every integer i, the edge v ; ->v i+1 is red if i mod 3 = 0 , white
if i mod 3 = 1 , and blue if i mod 3 = 2 .

Describe an algorithm to find the shortest French flag walks from one
starting vertex s to every other vertex in G.

18. There are n galaxies connected by m intergalactic teleport-ways. Each
teleport-way joins two galaxies and can be traversed in both directions. Also,
each teleport-way e has an associated cost of c(e) dollars, where c(e) is a
positive integer. A teleport-way can be used multiple times, but the toll must
be paid every time it is used.

300

Exercises

Judy wants to travel from galaxy s to galaxy t as cheaply as possible.
However, she wants the total cost to be a multiple of five dollars, because
carrying small change is not pleasant either.

(a) Describe and analyze an algorithm to compute the minimum total cost
of traveling from galaxy s to galaxy t, subject to the restriction that the
total cost is a multiple of five dollars.

(b) Solve part (a), but now assume that Judy has a coupon that allows her
to use exactly one teleport-way for free.

19. After moving to a new city, you decide to choose a walking route from your
home to your new office. To get a good daily workout, your route must
consist of an uphill path (for exercise) followed by a downhill path (to cool
down), or just an uphill path, or just a downhill path. (You’ll walk the same
path home, so you’ll get exercise one way or the other.) But you also want
the shortest path that satisfies these conditions, so that you actually get to
work on time.

Your input consists of an undirected graph G, whose vertices represent
intersections and whose edges represent road segments, along with a start
vertex s and a target vertex t. Every vertex v has an associated value h(v),
which is the height of that intersection above sea level, and each edge uv
has an associated value (.(uv), which is the length of that road segment.

(a) Describe and analyze an algorithm to find the shortest uphill-downhill
walk from s to t. Assume all vertex heights are distinct.

(b) Now suppose we allow some or all vertex heights to be equal. Describe
and analyze an algorithm to find the shortest “uphill then downhill” walk
from s to t; you may use flat edges in both the “uphill” and “downhill”
portions of your walk.

(c) Finally, suppose you discover that there is no path from s to t with the
structure you want. Describe an algorithm to find a path from s to t
that alternates between “uphill” and “downhill” subpaths as few times
as possible, and has minimum length among all such paths.

20. After graduating from Sham-Poobanana University you accept a job with
Aerophobes-JI-Us, the leading traveling agency for people who hate to fly.
Your job is to build a system to help customers plan airplane trips from one
city to another. All of your customers are afraid of flying (and by extension,
airports), so any trip you plan needs to be as short as possible. You know all
the departure and arrival times of all the flights on the planet.

Suppose one of your customers wants to fly from city X to city Y.
Describe an algorithm to find a sequence of flights that minimizes the total

301

8. Shortest Paths

time in transit —the length of time from the initial departure to the final
arrival, including time at intermediate airports waiting for connecting flights.

21. In Exercise 20 from Chapter 5, you designed an algorithm to decide whether
a given acute-angle maze is solvable. In this problem, you will design
algorithms to find the shortest walk through a given acute-angle maze, for
two different definitions of "length".

Complete each angle maze below by tracing a path from start to finish that
has only acute angles.

Start

Your input is a connected undirected graph G whose vertices are points
in the plane and whose edges are line segments. Edges do not intersect,
except at their endpoints. For example, a drawing of the letter X would
have five vertices and four edges; the first maze above has 13 vertices and 15
edges. You are also given two vertices Start and Finish.

A walk from Start to Finish in G is valid if it contains only acute angles,
or more formally, for any two consecutive edges u->v->w, either Zuvw = n
or 0 < Zuvw < n/ 2 . Assume you can determine in 0 ( 1 ) time whether the
angle between two given segments is straight, obtuse, right, or acute.

(a) Describe an algorithm to compute a valid walk from Start to Finish that
traverses as few segments as possible. (If your walk traverses the same
segment twice, count it twice.)

(b) Describe an algorithm to compute a valid walk from Start to Finish that
makes as few turns as possible. [Hint: This is not the same as part (a).]

(c) Describe an algorithm to compute a valid walk from Start to Finish whose
total Euclidean length is as small as possible. (Assume you can also
compute the length of any segment in 0(1) time.)

22. After a grueling midterm at the See-Bull Center for Fake News Detection,
you decide to take the bus home. Since you planned ahead, you have a
schedule that lists the times and locations of every stop of every bus in
Sham-Poobanana. Unfortunately, no single bus visits both the See-Bull
Center and your home; you must change buses at least once. There are
exactly b different buses. Each bus starts at 12:00:01AM, makes exactly n
stops, and finally stops running at 11:59:59PM. Buses always run exactly on
schedule, and you have an accurate watch. Finally, you are far too tired to
walk between bus stops.

302

Exercises

(a) Describe and analyze an algorithm to determine the sequence of bus
rides that gets you home as early as possible. Your goal is to minimize
your arrival time, not the time you spend traveling.

(b) Oh, no! The midterm was held on Halloween, and the streets are infested
with zombies! The Sham-Poobanana Mass Transit District doesn’t have
the funding to add additional buses or install zombie-proof bus stops,
especially for only one night a year. Describe and analyze an algorithm
to determine a sequence of bus rides that minimizes the total time you
spend waiting at bus stops; you don’t care how late you get home or
how much time you spend on buses. (Assume you can wait inside the
See-Bull Center until your first bus is just about to leave.)

23. The first morning after returning from a glorious spring break, Alice wakes
to discover that her car won’t start, so she has to get to her classes at
Sham-Poobanana University by public transit. She has a complete transit
schedule for Poobanana County. The bus routes are represented in the
schedule by a directed graph G, whose vertices represent bus stops and
whose edges represent bus routes between those stops. For each edge u^v,
the schedule records three positive real numbers:

• £(u->v) is the length of the bus ride from stop u to stop v (in minutes)

• /(u->v) is the first time (in minutes past 12am) that a bus leaves stop u
for stop v.

• A(u->v ) is the time between successive departures from stop u to stop v
(in minutes).

Thus, the first bus for this route leaves u at time f{u->v ) and arrives at v at
time/(u^v)+£(u->v), the second bus leaves u at time/(iz->v)+A(iz^v) and
arrives at v at time /(u->v)+A(iz^v)+£(u->v), the third bus leaves u at time
f(u->v) + 2 - A(u->v) and arrives at v at time f(u->v) + 2 -A(u->v) + (.(u^v),
and so on.

Alice wants to leaves from stop s (her home) at a certain time and
arrive at stop t (The See-Bull Center for Fake News Detection) as quickly as
possible. If Alice arrives at a stop on one bus at the exact time that another
bus is scheduled to leave, she can catch the second bus. Because she’s a
student at SPU, Alice can ride the bus for free, so she doesn’t care how many
times she has to change buses.

Describe and analyze an algorithm to find the earliest time Alice can
reach her destination. Your input consists of the directed graph G = (V, E),
the vertices s and t, the values £(e),/(e), A(e) for each edge e € E, and
Alice’s starting time (in minutes past 12am).

[Hint: In this rare instance, it may be easier to modify the algorithm,
instead of modifying the input graph.]

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8 . Shortest Paths

24. Mulder and Scully have computed, for every road in the United States,
the exact probability that someone driving on that road won’t be abducted
by aliens. Agent Mulder needs to drive from Langley, Virginia to Area 51,
Nevada. What route should he take so that he has the least chance of being
abducted?

More formally, you are given a directed graph G = (V, E), where every
edge e has an independent safety probability p(e). The safety of a path is
the product of the safety probabilities of its edges. Design and analyze an
algorithm to determine the safest path from a given start vertex s to a given
target vertex t. You may assume that all necessary arithmetic operations
can be performed in 0(1) time.

Las Vegas, NV

For example, with the probabilities shown above, if Mulder tries to drive
directly from Langley to Area 51, he has a 50 % chance of getting there
without being abducted. If he stops in Memphis, he has a 0.7 x 0.9 = 63 %
chance of arriving safely. If he stops first in Memphis and then in Las Vegas,
he has a 1 — 0.7 x 0.1 x 0.5 = 96 . 5 % chance of being abducted! (That’s how
they got Elvis, you know.)

*25. On an overnight camping trip in Sunnydale National Park, you are woken
from a restless sleep by a scream. As you crawl out of your tent to investigate,
a terrified park ranger runs out of the woods, covered in blood and clutching
a crumpled piece of paper to his chest. As he reaches your tent, he gasps,
“Get out. .. while... you... ”, thrusts the paper into your hands, and falls to
the ground. Checking his pulse, you discover that the ranger is stone dead.

You look down at the paper and recognize a map of the park, drawn
as an undirected graph, where vertices represent landmarks in the park,
and edges represent trails between those landmarks. (Trails start and end
at landmarks and do not cross.) You recognize one of the vertices as your
current location; several vertices on the boundary of the map are labeled
EXIT.

On closer examination, you notice that someone (perhaps the poor dead
park ranger) has written a real number between o and 1 next to each vertex

304

Exercises

and each edge. A scrawled note on the back of the map indicates that a
number next to an edge is the probability of encountering a vampire along
the corresponding trail, and a number next to a vertex is the probability of
encountering a vampire at the corresponding landmark. (Vampires can’t
stand each other’s company, so you’ll never see more than one vampire on
the same trail or at the same landmark.) The note warns you that stepping
off the marked trails will result in a slow and painful death.

You glance down at the corpse at your feet. Yes, his death certainly
looked painful. Wait, was that a twitch? Are his teeth getting longer? After
driving a tent stake through the undead ranger’s heart, you wisely decide to
immediately leave the park as fast as possible.

Describe and analyze an efficient algorithm to find a path from your
current location to an arbitrary EXIT node, such that the total expected
number of vampires encountered along the path is as small as possible. Be
sure to account for both the vertex probabilities and the edge probabilities.
[Hint: Even without the vertex probabilities, this is not the same as the
previous problem!]

305

The tree which fills the arms grew from the tiniest sprout;
the tower of nine storeys rose from a (small) heap of earth;
the journey of a thousand li commenced with a single step.

- Lao-Tzu, Tao Te Ching, chapter 64 (6th century BC),
translated by James Legge (1891)

And I would walk five hundred miles,

And I would walk five hundred more,

Just to be the man who walks a thousand miles
To fall down at your door.

- The Proclaimers, “I'm Gonna Be (500 Miles)",
Sunshine on Leith (2001)

Almost there... Almost there...

- Red Leader [Drewe Henley], Star Wars (1977)

9

All-Pairs Shortest Paths

g.i Introduction

In the previous chapter, we discussed several algorithms to find the shortest
paths from a single source vertex s to every other vertex of the graph, by
constructing a shortest path tree rooted at s. The shortest path tree specifies
two pieces of information for each node v in the graph:

• dist(v) is the length of the shortest path from s to v;

• pred(v) is the second-to-last vertex in the shortest path from s to v.

In this chapter, we consider the more general all pairs shortest path problem,
which asks for the shortest path from every possible source to every possible
destination. For every pair of vertices u and v, we want to compute the following
information:

dist(u, v) is the length of the shortest path from u to v;

pred{u, v) is the second-to-last vertex on the shortest path from u to v.

307

g. All-Pairs Shortest Paths

These intuitive definitions exclude a few boundary cases, all of which we
already saw in the previous chapter.

• If there is no path from u to v, then there is no shortest path from u to v; in
this case, we define dist(u, v) = oo and pred{u, v) = Null.

• If there is a negative cycle between u and v, then there are paths 1 from u to
v with arbitrarily negative length; in this case, we define dist(u, v) = —oo
and pred{u, v) = Null.

• Finally, if u does not lie on a negative cycle, then the shortest path from u to
itself has no edges, and therefore doesn’t have a last edge; in this case, we
define dist(u, u) = 0 and pred{u, u ) = Null

The desired output of the all-pairs shortest path problem is a pair of V x V arrays,
one storing all V 2 shortest-path distances, 2 the other storing all V 2 predecessors.
In this chapter, I’ll focus almost exclusively on computing the distance array.
The predecessor array, from which we can compute the actual shortest paths,
can be computed with only minor modifications (hint, hint).

g.2 Lots of Single Sources

The obvious solution to the all-pairs shortest path problem is to run a single¬
source shortest path algorithm V times, once for each possible source vertex.
Specifically, to fill in the one-dimensional subarray dist[s, •], we invoke a single¬
source algorithm starting at the source vertex s.

ObviousAPSP(V, E, w):
for every vertex s

dist[s,-] <— SSSP(V, E,w,s)

The running time of this algorithm obviously depends on which single-source
shortest path algorithm we use. Just as in the single-source setting, there are
four natural options, depending on the structure of the graph and its edge
weights:

• If the edges of the graph are unweighted, breadth-first search gives us an
overall running time of 0(VE ).

• If the graph is acyclic, scanning the vertices in topological order gives us an
overall running time of 0(VE ).

‘formally, walks

2 Back when road maps used to be printed on paper and had to be searched manually, it
was fairly common for them to include a triangular “distance table”. To find the distance from
Champaign to Columbus, for example, you would look in the row labeled “Champaign” and the
column labeled “Columbus”.

308

g.3- Reweighting

• If all edge weights are non-negative, Dijkstra’s algorithm gives us a running
time to 0 (VE log V) = 0 (V 3 log V). 3

• Finally, in the most general setting, the Bellman-Ford algorithm gives us an
overall running time of 0 (V 2 E) = 0 (V 4 ).

9.3 Reweighting

Negative edges slow us down; can we get rid of them? One simple idea that
occurs to many people is increasing the weights of all the edges by the same
amount so that all the weights become positive, so that we can use Dijkstra’s
algorithm instead of Bellman-Ford. Unfortunately, this simple idea doesn’t work,
intuitively because our two natural notions of “length” are incompatible—paths
with more edges can have smaller total weight than paths with fewer edges. If
we increase all edge weights at the same rate, paths with more edges get longer
faster than paths with fewer edges; as a result, the shortest path between two
vertices might change.

Figure 9.1. Increasing all the edge weights by 2 changes the shortest path froms to t.

However, there is a more subtle method for reweighting edges that does
preserve shortest paths. This reweighting method is often attributed to Donald
Johnson, who described its application to shortest path algorithms in 1973.
In fact, Johnson attributed the method to a t972 paper of Jack Edmonds and
Richard Karp; the same method was also described in a slightly different form
by Delbert Fulkerson in t96r.

Suppose each vertex v has some associated price n(v), which might be
positive, negative, or zero. We can define a new weight function w' as follows:

w'(u->v ) = 71(11) + w(u->v) — n(v)

To give some intuition, imagine that when we leave vertex u, we have to pay an
exit tax of tr(iz), and when we enter v, we get n(v) as an entrance gift.

It’s not hard to show that shortest paths with the new weight function w'
are exactly the same as shortest paths with the original weight function w. In

3 Again, if we replace the binary heap in our implementation of Dijkstra’s algorithm with

an unsorted array, the overall running time becomes 0 (V 3 ) (no matter how many edges the
graph has), and if we replace the binary heap with a Fibonacci heap, the running time drops to
0(V(E + V log V)) = 0(VE + V 2 log V) = 0(V 3 ).

309

g. All-Pairs Shortest Paths

fact, for any path u ~> v from one vertex u to another vertex v, we have
w'(u v) = 7t(iz) + w(u v) — 7t(v).

We pay 7t(u) in exit fees, plus the original weight of of the path, minus the 7t(v)
entrance gift. At every intermediate vertex x on the path, we get tt(x) as an
entrance gift, but then immediately pay it back as an exit tax! Since all paths
from u to v change length by exactly the same amount, the shortest path from u
to v does not change. (Paths between different pairs of vertices could change
lengths by different amounts, so their order could change.)

9.4 Johnson's Algorithm

Johnson’s all-pairs shortest path algorithm computes a cost tr(v) for each vertex,
so that the new weight of every edge is non-negative, and then computes shortest
paths with respect to the new weights using Dijkstra’s algorithm.

First, suppose the input graph has a vertex s that can reach all the other
vertices. Johnson’s algorithm computes the shortest paths from s to the other
vertices, using Bellman-Ford (which doesn’t care if the edge weights are nega¬
tive), and then reweights the graph using the price function tt(v) = dist(s,v).
The new weight of every edge is

w'(u->v) = dist(s, u ) + w(u->v) — dist(s, v).

These new weights non-negative because Bellman-Ford halted! Recall that
an edge u->v is tense if dist(s, u ) + w(u->v) < dist(s, v), and that single-source
shortest path algorithms eliminate all tense edges. (If Bellman-Ford detects
a negative cycle, Johnson’s algorithm aborts, because shortest paths are not
well-defined.)

If there is no suitable vertex s that can reach everything, then no matter
where we start Bellman-Ford, some of the resulting vertex prices will be infinite.
To avoid this issue, we always add a new vertex s to the graph, with zero-weight
edges from s to the other vertices, but no edges going back into s. This addition
doesn’t change the shortest paths between any pair of original vertices, because
there are no paths into s.

Complete pseudocode for Johnson’s algorithm is shown in Figure 9.2. The
running time of this algorithm is dominated by the calls to Dijkstra’s algorithm.
Specifically, we spend 0 (VE ) time running BellmanFord once, 0 {VE logV)
time running Dijkstra V times, and 0 (V + E ) time doing other bookkeeping.
Thus, the overall running time is O(VElogV) = 0 (V 3 logV). 4 Negative edges
don’t slow us down after all!

4 ... assuming the default binary-heap implementation; see the previous footnote.

310

g.5. Dynamic Programming

JohnsonAPSP(V,E, w) :

((Add an artificial source))
add a new vertex s
for every vertex v

add a new edge s->v
w(s^v) <— 0

((Compute vertices prices))
dist[s, •] «— BellmanFord( V,£,w,s)
if BellmanFord found a negative cycle
fail gracefully
((Reweight edges))
for every edge (u, v)e£

w'(u-*v ) <— dist[s, u] + w(u^v) — dist[s, v]

((Compute reweighted shortest paths))
for every vertex u

dist'[u, •] «— Dijkstra(V,£, w' ,u)

((Compute original shortest-path distances))
for every vertex u
for every vertex v

dist[u, v] <— dist'[u, v] — dist[s, u] + dist[s, v]

Figure 9.2. Johnson's all-pairs shortest paths algorithm

9.5 Dynamic Programming

We can also solve the all-pairs shortest path problem using dynamic program¬
ming, instead of invoking a single-source algorithm. For dense graphs, where
E = Q(V 2 ), the dynamic programming approach eventually yields an algorithm
that is both simpler and (slightly) faster than Johnson’s algorithm. For the rest
of this chapter, I will assume that the input graph contains no negative
cycles.

As usual for dynamic programming algorithms, we first need a recurrence.
Just as in the single-source setting, the “obvious” recursive definition

only works when the input graph is a dag; any directed cycles drive the
recurrence into an infinite loop.

We can break this infinite loop by introducing as an additional parameter,
exactly as we did for Bellman-Ford; let dist(u,v,t) denote the length of the
shortest path from u to v that uses at most l edges. The shortest path between
any two vertices traverses at most V — l edges, so the true shortest-path distance
is dist(u,v,V — 1 ). Bellman’s single-source recurrence adapts to this setting

311

g. All-Pairs Shortest Paths

immediately:

dist(u,v,l )

0

oo
min ■

dist(u, v, i — 1)

min ( dist(u,x,l — 1) + w(x->v))

X—»V

if l = 0 and u = v
if l = 0 and u^v

otherwise

Turning this recurrence into a dynamic programming algorithm is straightfor¬
ward; the resulting algorithm runs in 0 (V 2 E) = 0 (V 4 ) time.

ShimbelAPSP(V, E, w):

for all vertices u

for all vertices v

>

II

3

M-h

dist[u, v, 0] <— 0

else

dist[u, v, 0] <— 00

for l <— 1 to V — 1

for all vertices u

for all vertices v/u

dist[u, v, l ] <— dist[u

v,£-l]

for all edges x^v

if dist[u, v,£] >

dist[u, x, t — 1] 4- w(x^v)

dist[u,v,l ]

<— dist[u, x, i — 1 ] + w(x-w)

This algorithm was first sketched by Alfonso Shimbel in 1954. 5 Just like
Bellman’s formulation of Bellman-Ford, we don’t need the inner loop over
vertices v or the iteration index L The modified algorithm is shown below.

AllPairsBellmanFord(V, E, w):
for all vertices u
for all vertices v
if u = v

dist[u, v] <— 0

else

dist[u, v] <— 00

for £ <— 1 to V — 1
for all vertices u

for all edges x->v

if dist[u, v] > dist[u,x] + w(x^v)
dist[u, v] «— dist[u, x] + w(x->v)

5 Shimbel assumed the input was a complete V x V matrix of distances, so his original
algorithm actually runs in 0(V 4 ) time no matter how many edges the graph has.

312

g.6. Divide and Conquer

Given how we derived it, it should come as no surprise that the resulting
algorithm is exactly the same as interleaving V different executions of Bellman-
Ford, one running at each vertex. In particular, for all vertices u and v, after the
fth iteration of the main for-loop, dist[u, v] is at most the length of the shortest
path from u to v containing at most l edges.

g.6 Divide and Conquer

But we can make a more significant improvement, suggested by Michael Fischer
and Albert Meyer in 1971. Bellman’s recurrence breaks the shortest path into a
slightly shorter path and a single edge, by considering all possible predecessors
of the target vertex. Instead, let’s break the shortest paths into two shorter
shortest paths at the middle vertex. This idea gives us a different recurrence
for same the function dist(u, v, £). Here we need to stop at the base case 1 = 1
instead of l = 0, because a path with at most one edge has no “middle” vertex.
To simplify the recurrence slightly, let’s define w(v->v) = 0 for every vertex v.

{ w(u->v) if i = 1

, ,

min(dist(u, x,£/2) + dist{x, v,t/2 )J otherwise

As stated, this recurrence only works when l is a power of 2 , since otherwise we
might try to find the shortest path with (at most) a fractional number of edges!
But that’s not really a problem; dist{u,v,l) is the true shortest-path distance
from u to v for all 1 >V — 1 ; in particular, we can use l = 2 ^ v ^ < 2 V.

Once again, a dynamic programming solution is straightforward. Even before
we write down the algorithm, we can tell the running time is 0(V 3 log V) —we
need to consider V possible values of u, v, and x, but only |TgV] possible values
of l. In the following pseudocode for Fischer and Meyer’s algorithm, the array
entry dist[u, v, i] stores the value of dist(u, v, 2 ! ).

FischerMeyerAPSP(V,E, w):
for all vertices u
for all vertices v

dist[u, v, 0] <— w(ihv)

for i <— 1 to rigVl ((£ = 2'))

for all vertices u
for all vertices v

dist[u, v, i] <— 00
for all vertices x

if dist[u, v, i] > dist[u,x, i — 1] + dist[x, v, i — 1]
dist[u, v, i] <— dist[u,x, i — 1] + dist[x, v, i — 1]

Unlike our earlier algorithms, FischerMeyerAPSP is not the same as V
invocations of any single-source shortest-path algorithm; in particular, the

313

g. All-Pairs Shortest Paths

innermost loop does not simply relax tense edges. Nevertheless, we can still
remove the last dimension of the table, using dist[u, v] everywhere in place of
dist[u, v, i], just as we did in Bellman-Ford and our earlier dynamic programming
algorithm; this reduces the space from 0 (V 3 ) to 0 (V 2 ). This more polished
algorithm was described by Leyzorek et al. in t957, in the same paper where
they describe Dijkstra’s algorithm.

LeyzqrekAPSP(V, E, w):
for all vertices u
for all vertices v

dist[u, v] <— w(u—>v)

for i «— 1 to rigVl ((£ = 2 1 ))

for all vertices u
for all vertices v

for all vertices x

if dist[u, v] > dist[u, x] + dist[x, v]
dist[u, v] <— dist[u, x] + dist[x, v]

9.7 Funny Matrix Multiplication

There is a very close connection (first observed by Shimbel, and later indepen¬
dently by Bellman) between computing shortest paths in a directed graph and
computing powers of a square matrix. Compare the following algorithm for
squaring an nx n matrix A with the inner loop of FischerMeyerAPSP. (I’ve
slightly modified the notation in the second algorithm to make the similarity
clearer.)

MatrixSquare(A) :
for i <— 1 to n

for j <— 1 to n
A'[i,j ]«-0
for k *— 1 to n

A'[i,j ] «- A'[i,j] +A[i,k ] -A[k,j]

FischerMeyerInnerLoop(D) :
for all vertices u
for all vertices v
D'[u,v] <— 00
for all vertices x

D'[u,v] <— minjD'fu,v], D[u,x] +D[x,v]}

The only difference between these two algorithms is that the second algorithm
uses addition instead of multiplication, and minimization instead of addition. For
this reason, the shortest path inner loop is sometimes referred to as “min-plus”
or “distance” or “funny” matrix multiplication.

3M

97- Funny Matrix Multiplication

Our slower algorithm ShimbelAPSP is the standard iterative algorithm for
computing the (V — l)th “min-plus power” of the weight matrix w. The first
set of loops sets up the min-plus identity matrix, with Os on the main diagonal
and oo everywhere else, and each iteration of the second main loop computes
the next “min-plus power”. FischerMeyerAPSP replaces this iterative method
for computing powers with repeated squaring, exactly as we saw at the end of
Chapter 1. Once again, we see the influence of ancient Egyptian ciQJtsdovdTtxad.

There are faster divide-and-conquer algorithms for (standard) matrix multi¬
plication, similar to Karatsuba’s divide-and-conquer algorithm for multiplying
integers. The first such algorithm, described by Volker Strassen in 1969, reduces
the problem of multiplying two n x n matrices to seven instances of multiplying
two n/ 2 x n /2 matrices; Strassen’s algorithm runs in 0 (n lg7 ) = 0 (n 2 807355 ).
Strassen’s algorithm has been improved many times over the last fifty years; as
of 2018, the fastest matrix-multiplication algorithm known runs in 0(n 2 ' 372864 )
time. * * 6 Unfortunately, all of these faster algorithms use subtraction, and there’s
no “funny” equivalent of subtraction. (What’s the inverse operation for min?)
So at least for general graphs, there’s no obvious way to speed up the inner loop
of our dynamic programming algorithms.

But “not obvious” does not mean “impossible”! In fact, there are several
significantly faster algorithms for special cases of the all-pairs shortest paths
problem. One of the nicest is a simple randomized algorithm discovered in
1991 by Zvi Galil and Oded Margalit, and further simplified in 1992 by Raimund
Seidel, that computes all-pairs shortest path distances in unweighted, undirected
graphs in 0 (M(V) log V) expected time, where M(n) = 0 (n 2 372864 ) is the time
required to (seriously) multiply two nxn integer matrices. 7 Galil, Margalit, and
Seidel’s approach has since been extended to compute actual shortest paths,
deterministically, in directed graphs, with small integer edge weights, in strongly
subcubic time.

On the other hand, despite considerable progress in the small-integer-weight
setting, nobody knows how to compute all-pairs shortest paths for more general
edge weights in 0 (V 2 ' 999999 ) time, for any number of 9 s. Moreover, there is
some evidence that such an algorithm is impossible! So maybe “not obvious”
does mean “impossible” after all.

determining the minimum time required to multiply two arbitrary nxn matrices is a

long-standing open problem; many people believe there is an undiscovered algorithm that runs

in 0(n 2+e ) time for any e > 0, or possibly even in 0(n 2 ) time.

7 Raimund Seidel. On the all-pairs-shortest-path problem in unweighted undirected graphs.
Journal of Computer and System Sciences, 51(31:400-403,1995. This is one of the few algorithms
papers where (in the 1992 conference version at least) the algorithm is completely described
and analyzed in the abstract of the paper. See also: Noga Alon, Zvi Galil, Oded Margalit*. On
the exponent of the all pairs shortest path problem. Journal of Computer and System Sciences
54(23:255-262,1997.

3d

g. All-Pairs Shortest Paths

9.8 (Kleene-Roy-)Floyd-Warshall(-lngerman)

Our fast dynamic programming algorithm is still a factor of O(logV) slower
in the worst case than the standard implementation of Johnson’s algorithm.
A different formulation of shortest paths that removes this logarithmic factor
was proposed twice in 1962, first by Robert Floyd and later independently by
Peter Ingerman, both slightly generalizing an algorithm of Stephen Warshall
published earlier in the same year. In fact, Warshall’s algorithm was previously
discovered by Bernard Roy in 1959, and the underlying recursion pattern was
used by Stephen Kleene 8 in 1951.

Warshall’s (and Roy’s and Kleene’s) insight was to use a different third
parameter in the dynamic programming recurrence. Instead of considering
paths with a limited number of edges, they considered paths that can pass
through only certain vertices. Here, “pass through” means “both enter and
leave”; for example, the path w^x^y^z starts at w, passes through x and y,
and ends at z.

Number the vertices arbitrarily from 1 to V. For every pair of vertices u
and v and every integer r, we define a path n{u, v, r) as follows:

n{u, v, r) is the shortest path (if any) from u to v that passes
through only vertices numbered at most r.

In particular, n{u, v, V) is the true shortest path from u to v. Kleene and Roy
and Warshall all observed that these paths have a simple recursive structure.

intermediate nodes < r-1

Figure 9.3. Recursive structure of the restricted shortest path n[u, v, r)

• The path n(u, v, 0 ) can’t pass through any intermediate vertices, so it must
be the edge (if any) from u to v.

• For any integer r > 0 , either n{u, v, r) passes through vertex r or it doesn’t.

- If tt(iz, v, r) passes through vertex r, it consists of a subpath from u to r,
followed by a subpath from r to v. Both of those subpaths pass through
only vertices numbered at most r — 1; moreover, those subpaths are as
short as possible with this restriction. So the two subpaths must be
n{u, r, r — 1) and n(r, v, r — 1).

8 Pronounced “clay knee”, not “clean” or “clean-ee” or “clay-nuh” or “dimaggio”. Specifically,
Kleene described an inductive proof that every finite automata has an equivalent regular expres¬
sion; Kleene’s induction pattern is essentially identical to the Floyd-Warshall recurrence.

316

g.8. (Kleene-Roy-)Floyd-Warshall(-lngerman)

- On the other hand, if n{u, v, r) does not pass through vertex r, then it
passes through only vertices numbered at most r — 1, and it must be
the shortest path with this restriction. So in this case, we must have
n(u, v, r) = n{u, v,r — 1).

Now let dist(u, v, r) denote the length of the path n{u, v, r). The recursive
structure of n{u, v, r) immediately implies the following recurrence:

{ w(iz^v) if r = 0

f dist(u, v, r — 1) )

min < > otherwise

[dist(u, r, r — 1) + dist(r, v, r — 1) J

Our goal is to compute dist(u,v,V ) for all vertices u and v. Once again,
this recurrence can be evaluated by a straightforward dynamic programming
algorithm in 0(V 3 ) time.

KleeneAPSP(V,£,w):
for all vertices u
for all vertices v

dist[u, v, 0] <— w(u-tv)

for r «— 1 to V

for all vertices u
for all vertices v

if dist[u, v, r — 1] < dist[u, r, r — 1] + dist[r, v, r — 1]
dist[u, v, r ] <— dist[u, v, r — 1]

else

dist[u, v, r]«— dist[u, r, r — 1] + dist[r, v, r — 1]

Like all our previous dynamic programming algorithms for shortest paths, we
can simplify KleeneAPSP by removing the third dimension of the memoization
table. Also, because we chose the vertex numbering arbitrarily, there’s no
reason to refer to it explicitly in the pseudocode. We finally arrive at Floyd’s
improvement of Warshall’s algorithm:

FloydWarshall(V~, E, w):
for all vertices u
for all vertices v

dist[u, v] <— w(u-s-v)

for all vertices r

for all vertices u
for all vertices v

if dist[u, v] > dist[u, r] + dist[r, v]
dist[u, v] «— dist[u, r ] + dist[r, v]

It’s interesting to compare FloydWarshall with our earlier, slightly slower
dynamic programming algorithm LeyzorekAPSP. Instead of 0 (log V) passes

317

g. All-Pairs Shortest Paths

through all triples of vertices, FloydWarshall requires only a single pass, but
only because it uses a different nesting order for the three loops!

Exercises

1. (a) Describe a modification of LeyzorekAPSP that returns an array of

predecessor pointers, in addition to the array of shortest path distances,
still in 0 (V 3 logV) time.

(b) Describe a modification of FloydWarshall that returns an array of
predecessor pointers, in addition to the array of shortest path distances,
still in 0(V 3 ) time.

2. All of the algorithms discussed in this chapter fail if the graph contains a
negative cycle. Johnson’s algorithm detects the negative cycle in the initial¬
ization phase (via Bellman-Ford) and aborts; the dynamic programming
algorithms just return incorrect results. However, all of these algorithms can
be modified to return correct shortest-path distances, even in the presence
of negative cycles. Specifically, for all vertices u and v:

• If u cannot reach v, the algorithm should return dist[u, v] = oo.

• If u can reach a negative cycle that can reach v, the algorithm should
return dist[u, v] = —oo.

• Otherwise, there is a shortest path from u to v, so the algorithm should
return its length.

(a) Describe how to modify Johnson’s algorithm to return the correct
shortest-path distances, even if the graph has negative cycles.

(b) Describe how to modify LeyzorekAPSP to return the correct shortest-
path distances, even if the graph has negative cycles.

(c) Describe how to modify Floyd-Warshall to return the correct shortest-path
distances, even if the graph has negative cycles.

3. The algorithms described in this chapter can also be modified to return an
explicit description of some negative cycle in the input graph G, if one exists,
instead of only reporting whether or not G contains a negative cycle.

(a) Describe how to modify Johnson’s algorithm to return either the array
of all shortest-path distances or a negative cycle.

(b) Describe how to modify LeyzorekAPSP to return either the array of all
shortest-path distances or a negative cycle.

(c) Describe how to modify Floyd-Warshall to return either the array of all
shortest-path distances or a negative cycle.

318

Exercises

In all cases, if the input graph contains more than one negative cycle, your
algorithms may choose one arbitrarily.

4. Let G = (V, E ) be a directed graph with weighted edges; edge weights can
be positive, negative, or zero, but there are no negative cycles.

(a) Describe an efficient algorithm that either finds a cycle of length zero
in G, or correctly reports that no such cycle exists.

(b) Describe an efficient algorithm that constructs a subgraph H of G with
the following properties:

• Every vertex of G is a vertex of H.

• Every directed cycle in H has length 0 .

• Every directed cycle of length 0 in G is also a cycle in H.

In particular, if there are no zero-cycles in G, then H has no edges.

5. Let G = (V, E ) be a directed graph with weighted edges; edge weights can
be positive, negative, or zero. Suppose the vertices of G are partitioned
into k disjoint subsets V 1 , V 2 ,..., V k ; that is, every vertex of G belongs to
exactly one subset V,. For each i and j, let < 5 (i, j) denote the minimum
shortest-path distance between vertices in V t and vertices in Vj:

5 (i, j) = min {dist[v t , Vj) | V; € V; and Vj € Vj} .

Describe an algorithm to compute 5 (i,;) for all i and j. For full credit, your
algorithm should run in 0 (VE + kV log V) time.

6. In this problem we will discover how you, yes you, can be employed by
Wall Street and cause a major economic collapse! The arbitrage business
is a money-making scheme that takes advantage of differences in currency
exchange. In particular, suppose 1 US dollar buys 120 Japanese yen; 1 yen
buys 0.01 euros; and 1 euro buys 1.2 US dollars. Then, a trader starting with
$1 can convert their money from dollars to yen, then from yen to euros,
and finally from euros back to dollars, ending with $1.44! The cycle of
currencies $—>¥—>€ —> $ is called an arbitrage cycle. Of course, finding
and exploiting arbitrage cycles before the prices are corrected requires
extremely fast algorithms.

Suppose n different currencies are traded in your currency market. You
are given the matrix ff[l.. n, 1.. n] of exchange rates between every pair of
currencies; for each i and j, one unit of currency i can be traded for R[i, j]
units of currency;. (Do not assume that • .R[j, i] = 1 .)

(a) Describe an algorithm that returns an array V [1 .. n], where V[i] is the
maximum amount of currency i that you can obtain by trading, starting
with one unit of currency 1, assuming there are no arbitrage cycles.

319

g. All-Pairs Shortest Paths

(b) Describe an algorithm to determine whether the given matrix of currency
exchange rates creates an arbitrage cycle.

(c) Modify your algorithm from part (b) to actually return an arbitrage
cycle, if it exists.

7. Morty needs to retrieve a stabilized plumbus from the Clackspire Labyrinth.
He must enter the labyrinth using Rick’s interdimensional portal gun, traverse
the Labyrinth to a plumbus, then take that plumbus through the Labyrinth
to a fleeb to be stabilized, and finally take the stabilized plumbus back
to the original portal to return home. Plumbuses are stabilized by fleeb
juice, which any fleeb will release immediately after being removed from its
fleebhole. An unstabilized plumbus will explode if it is carried more than
137 flinks from its original storage unit. The Clackspire Labyrinth smells like
farts, so Morty wants to spend as little time there as possible.

Rick has given Morty a detailed map of the Clackspire Labyrinth, which
consist of a directed graph G = (V,E) with non-negative edge weights
(indicating distance in flinks), along with two disjoint subsets P CV and
F c V, indicating the plumbus storage units and fleebholes, respectively.
Morty needs to identify a start vertex s, a plumbus storage unit p € P, and a
fleebhole / e F, such that the shortest-path distance from p to / is at most
137 flinks long, and the length of the shortest walk -~>s is as short as

possible.

Describe and analyze an algo (burp) rithm to so(burp)olve Morty’s prob¬
lem. You can assume that it is in fact possible for Morty to succeed.

8. Let G = (V, E ) be a directed graph with weighted edges; edge weights could
be positive, negative, or zero.

(a) How would we delete an arbitrary vertex v from this graph, without
changing the shortest-path distance between any other pair of vertices?
Describe an algorithm that constructs a directed graph G' = (V \ {v},E')
with weighted edges, such that the shortest-path distance between any
two vertices in G' is equal to the shortest-path distance between the
same two vertices in G, in 0 (V 2 ) time.

(b) Now suppose we have already computed all shortest-path distances in G'.
Describe an algorithm to compute the shortest-path distances in the
original graph G from v to every other vertex, and from every other
vertex to v, all in 0 (V 2 ) time.

(c) Combine parts (a) and (b) into another all-pairs shortest path algorithm
that runs in 0 (V 3 ) time. (The resulting algorithm is almost the same as
Floyd-Warshall!)

320

Exercises

9. A third type of matrix multiplication is also relevant for shortest-path
algorithms. Suppose A and B are boolean n x n matrices. The boolean or
and-or product of A and B is the nxn matrix C where C[i, j] = \J k (A[i, k ] A
B[k,jJ).

(a) Reduce boolean matrix multiplication to min-plus matrix multiplication.
That is, given a subroutine MinPlusMultiply that computes the min-
plus product of two nxn matrices, describe and analyze an algorithm
BooleanMatrixMultiply that multiplies two boolean matrices.

(b) Reduce boolean matrix multiplication to standard matrix multiplication.
That is, given a subroutine MatrixMultiply that computes the standard
product of two nxn matrices, describe and analyze an algorithm
BooleanMatrixMultiply that multiplies two boolean matrices.

10. The transitive closure of a directed graph G contains an edge u->v if and
only if there is a directed path from u to v in G.

(a) Suppose we can multiply two nxn boolean matrices in 0 (n") time,
for some constant 2 < to < 3 . (Problem 9(b) implies co < 2 . 372864 .)
Describe an algorithm to compute the transitive closure of an n-vertex
directed graph in 0(n" log n) time.

(b) Now suppose G is a directed acyclic graph. Describe an algorithm to
compute the transitive closure of G in 0 (n") time. [Hint: Do what you
always do with dags, and then divide and conquer. Use the fact that
co >2.]

(c) Finally, describe an algorithm to compute the transitive closure of an
arbitrary directed graph in 0 (n") time. [Hint: Do what you always do
to turn an arbitrary directed graph into a dag.]

(d) Now let’s reverse the previous reduction. Given a subroutine Transitive-
Closure that computes the transitive closure of an n-vertex directed
graph in 0 (n“) time, for some constant 2 < a < 3 , describe and analyze
an algorithm for boolean matrix multiplication that runs in 0(n“) time.

11. Prove that the following recursive algorithm correctly computes all-pairs
shortest-path distances in 0 (n 3 ) time. For simplicity, you may assume n
is a power of 2 . As usual, the array D is passed by reference to the helper
function RecAPSP. [Hint: This is a jumbled version of Floyd-Warshall, with
significantly better cache behavior. 9 ]

9 Joon-Sang Park, Michael Penner, and Viktor K. Prasanna. Optimizing graph algorithms for
improved cache performance. IEEE Trans. Parallel and Distributed Systems 15(91:769-782, 2004.
For a significant generalization to a wider class of dynamic programming problems, see: Rezaul
Alam Chowdhury and Vijaya Ramachandran. Cache-oblivious dynamic programming. Proc. 17th
SODA 591-600, 2006.

321

g. All-Pairs Shortest Paths

RecursiveAPSPI V, E, w):

RecAPSP(D, n , t,fc):

n <— |Vj

if n= 1

for i <— 1 to n

D[i,j ]«- min {P[i, j], D[i, fc] + D[j, fc]}

for j *— 1 to n

else

if i = j

m <— n /2

D\i il <- 0

RecAPSP(D, n/ 2 , t, j, k )

RecAPSP(D, n/ 2 , i, j, k + m )

RecAPSP(D, n/ 2 , i, j + m, fc )

RecAPSP(D, n/ 2 , i, j + m, k + m)

D[i,j] <— 00

RecAPSP(D, n/ 2 , i + m, j, fc )

RecAPSP(D, n/ 2 , i + m, j, k + m )

REcAPSP(D,n, 1 , 1 , 1 )

RecAPSP(D, n/ 2 , i + m, j + m, fc )

return D[1.. n, 1.. n]

RecAPSP(D, n/ 2 , i + m, j + m, k + m)

*12. Let G = (V, E) be an undirected, unweighted, connected, n-vertex graph,
represented by the adjacency matrix A [1 .. n, 1 .. n]. In this problem, we will
derive Seidel’s sub-cubic algorithm to compute the nxn matrix D [ 1 .. n, 1 .. n]
of shortest-path distances in G using fast matrix multiplication. Assume that
we have a subroutine MatrixMultiply that computes the standard product
of two nxn matrices in 0(n 6J ) time, for some unknown constant co> 2.

(a) Let G 2 denote the graph with the same vertices as G, where two vertices
are connected by a edge if and only if they are connected by a path of
length at most 2 in G. Describe an algorithm to compute the adjacency
matrix of G 2 using a single call to MatrixMultiply and 0(n 2 ) additional
time.

(b) Suppose we discover that G 2 is a complete graph. Describe an algorithm
to compute the matrix D of shortest path distances in G in 0 (n 2 )
additional time.

(c) Suppose we recursively compute the matrix D 2 of shortest-path distances
in G 2 . Prove that the shortest-path distance in G from node i to node j
is either 2 • D 2 [i,j ] or 2 • D 2 [i,j ] — 1 .

(d) Now suppose G 2 is not a complete graph. LetX = D 2 - A, and let deg(i)
denote the degree of vertex i in the original graph G. Prove that the
shortest-path distance from node i to node j in G is 2 • D 2 [i, j] if and
only if X[i,j] > D 2 [i, j] • deg(i).

(e) Describe an algorithm to compute the matrix D of shortest-path distances
in G in 0 (n"logn) time.

13. Gideon Yuval proposed the following reduction from min-plus matrix mul¬
tiplication to standard matrix multiplication in 1976. Suppose we are
given two nxn matrices A and B of integers, all between 0 and M, and
we want to compute their min-sum product matrix C, defined by setting

322

Exercises

C[i, k ] = min(A[i, j] + B[j, k ]). Define two new n x n matrices A! and B',

where 1

A'[i,j] = n M ~ A[i ’i ] and B'[i,j] = n M “ B[i J' ] .

Finally, let C 7 be the (standard) product of A' and B', defined by setting

C'[i,k] = Z J A'[i,j]-B'[j,k].

(a) Describe an algorithm to construct A' from A using only standard arith¬
metic operations (+, —, x).

(b) Describe an algorithm to extract the min-sum product C from C 7 , using
only standard arithmetic operations (+, —, x). 10

(c) Suppose we can (seriously) multiply two n x n integer matrices using
0 (n") arithmetic operations, for some constant 2 < to < 3 . How
many arithmetic operations does Yuval’s algorithm need to compute the
min-sum product C?

(d) Given a single n x n integer matrix A, how many arithmetic operations are
required to compute the nth “funny” power of A using Yuval’s algorithm?
(Recall that if A is the weighted adjacency matrix of a graph, then the
nth “funny” power of A is the matrix of shortest-path distances.)

(e) Why doesn’t Yuval’s algorithm imply an all-pairs shortest path algorithm
that is faster than Floyd-Warshall for arbitrary edge weights? How are
we cheating?

“In particular, do not use division or the floor function [xj. Trust me—this is a can of worms
you do not want to open.

323

A process cannot be understood by stopping it. Understanding must move with
the flow of the process, must join it and flow with it.

— The First Law of Mentat, in Frank Flerbert's Dune (1965)

Contrary to expectation, flow usually happens not during relaxing moments of
leisure and entertainment, but rather when we are actively involved in a difficult
enterprise, in a task that stretches our mental and physical abilities.... Flow is hard
to achieve without effort. Flow is not "wasting time."

— Mihaly Csikszentmihalyi, Flow: The Psychology of Optimal Experience (1990)

There's a difference between knowing the path and walking the path.

— Morpheus [Laurence Fishburne], The Matrix (1999)

Maximum Flows & Minimum Cuts

In the mid-1950s, U. S. Air Force researcher Theodore E. Harris and retired U. S.
Army general Frank S. Ross wrote a classified report studying the rail network
that linked the Soviet Union to its satellite countries in Eastern Europe. The
network was modeled as a graph with 44 vertices, representing geographic
regions, and 105 edges, representing links between those regions in the rail
network. Each edge was given a weight, representing the rate at which material
could be shipped from one region to the next. Essentially by trial and error, they
determined both the maximum amount of stuff that could be moved from Russia
into Europe, as well as the cheapest way to disrupt the network by removing
links (or in less abstract terms, blowing up train tracks), which they called
“the bottleneck”. Their report, which included the drawing of the network in
Figure 10.1, was only declassified in 1999. 1

T first learned this story from Alexander Schrijver’s fascinating survey “On the history of
combinatorial optimization (till i960)”; the Harris-Ross report was declassified at Schrijver’s
request. Ford and Fulkerson (who we will meet shortly) credit Harris for formulating the

325

io. Maximum Flows & Minimum Cuts

Figure 10.1. Harris and Ross's map of the Warsaw Pact rail network. (See Image Credits at the end of
the book.)

This one of the first recorded applications of the maximum flow and
minimum cut problems. For both problems, the input is a directed graph
G = (V, E) with two special vertices s and t, called the source and target. As in
previous chapters, I will write u -^v to denote the directed edge from vertex u
to vertex v. Intuitively, the maximum flow problem asks for the maximum rate
at which some resource can be moved from sto t; the minimum cut problem
asks for the minimum damage needed to separate s from t.

10.1 Flows

An (s, t)-flow (or just a flow if the source and target vertices are clear from con¬
text) is a function / : £ —> R that satisfies the following conservation constraint
at every vertex v except possibly s and t:

2>( u—>v) = X>-w).

U W

In English, the total flow into v is equal to the total flow out of v. To keep the
notation simple, we define /(u->v) = 0 if there is no edge u->v in the graph.
The value of the flow /, denoted |/|, is the total net flow out of the source
vertex s:

I/I '-=^]f - (u^s).

326

maximum-flow problem, although the precise chronology is somewhat muddled; Harris and Ross
thank George Dantzig “for assistance in formulating the problem”.

10 . 2 . Cuts

It’s not hard to prove that |/1 is also equal to the total net flow into the target
vertex t, as follows. To simplify notation, let 3f(v) denote the total net flow
out of any vertex v:

0 /(v):=2/( u—>v)

U W

The conservation constraint implies that < 5 /(v) = 0 or every vertex v except s
and t, so

Y J df{v) = df(s) + df(t).

V

On the other hand, any flow that leaves one vertex must enter another vertex, so
we must have d/(v) = 0- It follows immediately that |/| = df(s ) = — 3 /(t).

Now suppose we have another function c: E — * R> 0 that assigns a non¬
negative capacity c(e) to each edge e. We say that a flow / is feasible (with
respect to c) if 0 < /(e) < c(e) for every edge e. Most of the time we consider
only flows that are feasible with respect to some fixed capacity function c. We say
that a flow / saturates edge e if /(e) = c(e), and avoids edge e if /(e) = 0 . The
maximum flow problem is to compute a feasible (s, t)-flow in a given directed
graph, with a given capacity function, whose value is as large as possible.

Figure 10.2. A feasible (s, t)-flow with value 10. Each edge is labeled with its flow/capacity.

10.2 Cuts

An (s, t)-cut (or just cut if the source and target vertices are clear from context)
is a partition of the vertices into disjoint subsets S and T —meaning S U T = V
and S n T = 0 —where s e S and t e T.

If we have a capacity function c: E —» M> 0 , the capacity of a cut is the sum
of the capacities of the edges that start in S and end in T:

l|S,r|| :=^^]c(v->w).

veS weT

(Again, if v->w is not an edge in the graph, we assume c(v->w) = 0 .) Notice that
the definition is asymmetric; edges that start in T and end in S are unimportant.

327

io. Maximum Flows & Minimum Cuts

The minimum cut problem is to compute an (s, t)-cut whose capacity is as
large as possible.

Figure 10.3. An (s, t)-cut with capacity 15. Each edge is labeled with its capacity.

Intuitively, the minimum cut is the cheapest way to disrupt all flow from s
to t. Indeed, it is not hard to show the following relationship between flows
and cuts:

Lemma 10.1. Let f be any feasible (s, t)-flow, and let ( S,T ) be any (s, t)-cut.
The value off is at most the capacity of(S, T ). Moreover, \f \ = ||S, T || if and
only iff saturates every edge from S to T and avoids every edge from T to S.

Proof: Choose your favorite flow / and your favorite cut (S, T), and then follow
the bouncing inequalities:

I/I = <3/(s) [by definition]

veS

[conservation constraint]

= SS /(v ^ w) “

veS w ves u

[math, definition of d ]

=Z2>-

veS w<£S

veS uf£S

[removing edges from S to S]

=Z2>-

vGS wgT

-*w) ~

veS ueT

[definition of cut]

vgS weT

->w)

[because /(u->v) > 0]

vgS WGT

>w)

[because /(v-»w) < c(v-»w)]

= ||S, T[| [by definition]

In the second step, we are just adding zeros, because < 3 /(v) = 0 for every vertex
v € S \ {s}. In the fourth step, we are removing flow values /(x-ry) where

328

10.3- The Maxflow-Mincut Theorem

both x and y are in S, because they appear in both sums: positively when v = x
and w = y, and negatively when v = y and u = x.

The first inequalities in this derivation is actually an equality if and only if /
avoids every edge from T to S. Similarly, the second inequality is actually an
equality if and only if / saturates every edge from S to T. □

This lemma immediately implies that if |/| = ||S, T||, then / must be a
maximum flow, and (S, T) must be a minimum cut.

10.3 The Maxflow-Mincut Theorem

Surprisingly, in every flow network, there is a feasible (s, t)-flow / and an (s, t)-
cut (S, T) such that |/| = ||S, T ||. This is the famous Maxflow-Mincut Theorem,
first proved by Lester Ford (of shortest-path fame) and Delbert Fulkerson in
1954 and independently by Peter Elias, Amiel Feinstein, and Claude Shannon
(of information-theory and maze-solving-robot fame) in 1956.

The Maxflow Mincut Theorem. In every flow network with source s and
target t, the value of the maximum (s, t)-flow is equal to the capacity of
the minimum (s, t)-cut.

Ford and Fulkerson proved this theorem as follows. Fix a graph G, vertices s
and t, and a capacity function c: E —* R> 0 . The proof will be easier if we assume
that G is reduced, meaning there is at most one edge between any two vertices u
and v. In particular, either c(u->v) = 0 or c(v->u) = 0 . This assumption is easy
to enforce: Subdivide each edge u->v in G with a new vertex x, replacing u->v
with a path u->x->v, and define c(u->x) = c(x->v) = c(u^v). The modified
graph has the same maximum flow value and minimum cut capacity as the
original graph.

Figure 10.4. Enforcing the one-direction assumption.

Let / be an arbitrary feasible (s, t)-flow in G. We define a new capacity
function Cf : V x V —* R, called the residual capacity, as follows:

r

Cf(u^v) = <

c(u^v) —/(u^v)
/(V-U)

0

if u^v € E
if v—>u G E
otherwise

329

io. Maximum Flows & Minimum Cuts

Intuitively, the residual capacity of an edge indicates how much more flow can be
pushed through that edge. Because f > 0 and / < c, these residual capacities
are always non-negative. It is possible to have Cf(u->v) > 0 even if u->v is not an
edge in the original graph G. Thus, we define the residual graph Gf = ( [V,Ef ),
where Ef is the set of edges whose residual capacity is positive. Most residual
graphs are not reduced; in particular, if 0 < /(u->v) < c(u->v), then the residual
graph Gf contains both tz->v and its reversal v

Figure 10.5. A flow / in a weighted graph G and the corresponding residual graph G f .

Now we have two cases to consider: Either there is a directed path from the
source vertex s to the target vertex t in the residual graph Gf , or there isn’t.

First suppose the residual graph Gf contains a directed path P fronts to t; we
call P an augmenting path. Let F = min u ,. veP Cf(u->v ) denote the maximum
amount of flow that we can push through P. We define a new flow f ': E —* R
(in the original graph) as follows:

{ /(u^v) + F ifu^veP
/(u^v) —F if v^iz e P
/(u->v) otherwise

5/10

Figure 10.6. An augmenting path with value F = 5 and the resulting augmented flow/'.

I claim that this new flow/ 7 is feasible with respect to the original capacities c,
meaning f> 0 and f <c everywhere. Consider an edge u->v in the original
graph G. There are three cases to consider.

• If the augmenting path P contains u->v, then

= f(u->v) + F > f(u->v) > 0

330

10.3- The Maxflow-Mincut Theorem

because / is feasible, and

/= f (u-»v) + F by definition of / 1

< f(u->v) + Cf(u->v) by definition of F

= /(u-»v) + c(u->v) — f (u->v) by definition of Cf
= c(u->v) Duh.

• If the augmenting path P contains the reversed edge v-m, then

//u-^v) = f(u->v) — F < f(u->v) < c(u->v),

again because / is feasible, and

/ / (u->v) = /(u->v) — F by definition of f'

> / (u->v) — Cy(v-iu) by definition of F

= / (tz->v) — / (u-m/) by definition of

= 0 Duh.

• Finally, if neither u^v nor v-^u is in the augmenting path, then /'(u-^v) =

f(u->v), and therefore 0 < < c(u^v), because / is feasible.

So / is indeed feasible.

Finally, only the first edge in the augmenting path leaves s, which implies
| f'\ = |/| + F > |/|. Thus, f' is a feasible flow with larger value than /. We
conclude that if there is a path from s to t in the residual graph Gf, then / is
not a maximum flow.

On the other hand, suppose the residual graph Gf does not contain a directed
path from s to t. Let S be the set of vertices that are reachable from s in Gf,
and let T = V \ S. The partition (S, T) is clearly an (s, t)-cut. For every vertex
u e S and v € T, we have

Cf{u^v) = (c(u-s-v)— f(u->-v)) + f(y-ni) = 0.

The feasibility of / implies c(u->v)—f(u->v) > 0 and f(v->u) > 0 , so in fact we
must have c(u^v) — f(u->v) = 0 and /(v->u) = 0 . In other words, our flow /
saturates every edge from S to T and avoids every edge from T to S. Lemma
lo.i now implies that |/| = ||S, T||, which means / is a maximum flow and (S, T)
is a minimum cut.

This completes the proof! □

331

io. Maximum Flows & Minimum Cuts

10.4 Ford and Fulkerson's augmenting-path algorithm

Ford and Fulkerson’s proof of the Maxflow-Mincut Theorem immediately sug¬
gests an algorithm to compute maximum flows: Starting with the zero flow,
repeatedly augment the flow along any path from s to t in the residual graph,
until there is no such path.

This algorithm has an important but straightforward corollary:

Integrality Theorem. If all capacities in a flow network are integers, then
there is a maximum flow such that the flow through every edge is an integer.

Proof: We argue by induction that after each iteration of the augmenting path
algorithm, all flow values and all residual capacities are integers.

• Before the first iteration, all flow values are 0 (which is an integer), and
all residual capacities are the original capacities, which are integers by
definition.

• In each later iteration, the induction hypothesis implies that the capacity F
of the augmenting path is an integer, so augmenting changes the flow on
each edge, and therefore the residual capacity of each edge, by an integer.

In particular, each iteration of the augmenting path algorithm increases the
value of the flow by a positive integer. It follows that the algorithm eventually
halts and returns a maximum flow. □

If every edge capacity is an integer, then conservatively, the Ford-Fulkerson
algorithm halts after at most |/*| iterations, where /* is the actual maximum
flow. In each iteration, we can build the residual graph G ( - and perform a
whatever-first-search to find an augmenting path in 0 (E) time. Thus, in this
setting, the algorithm runs in 0(E|/*|) time in the worst case.

The following example shows that this running time analysis is essentially
tight. Consider the 4-node network illustrated below, where X is some large inte¬
ger. The maximum flow in this network is clearly 2 X. However, Ford-Fulkerson
might alternate between pushing one unit of flow along the augmenting path
s—>iz—»v—>t and then pushing one unit of flow along the augmenting path
s—>v—>t, leading to a running time of 0 (X) = Q(|/*|).

Figure 10.7. A bad example for the Ford-Fulkerson algorithm

332

104- Ford and Fulkerson's augmenting-path algorithm

Ford and Fulkerson’s algorithm works quite well in many practical situations,
or in settings where the maximum flow value I/*! is small, but without further
constraints on the augmenting paths, this is not an efficient algorithm in worst
case. The example network above can be described using only O(logX) bits;
thus, the running time of Ford-Fulkerson is actually exponential in the input
size.

^Irrational Capacities

But what if the capacities are not integers? If we multiply all the capacities by
the same (positive) constant, the maximum flow increases everywhere by the
same constant factor. It follows that if all the edge capacities are rational, then
the Ford-Fulkerson algorithm eventually halts, although still in exponential time
(in the number of bits used to described the input).

However, if we allow irrational capacities, the algorithm can actually loop
forever, always finding smaller and smaller augmenting paths. Worse yet, this
infinite sequence of augmentations may not even converge to the maximum flow,
or even to a significant fraction of the maximum flow! The smallest network
that exhibits this bad behavior was discovered by Uri Zwick in 1993. 2

Consider the six-node network shown in Figure 10 . 8 . Six of the nine edges
have some large integer capacity X, two have capacity 1, and one has capacity
cf) = (a /5 — l )/2 ^ 0.618034, chosen so that 1 — </> = </> 2 . To prove that the
Ford-Fulkerson algorithm can get stuck, we can watch the residual capacities of
the three horizontal edges as the algorithm progresses. (The residual capacities
of the other six edges will always be at least X — 3.)

Figure 10.8. Uri Zwick's non-terminating flow example, and three augmenting paths.

2 Ford and Fulkerson described a network with the same bad behavior in 1962, which had 10
vertices and 48 edges.

333

io. Maximum Flows & Minimum Cuts

Suppose the Ford-Fulkerson algorithm starts by choosing the central aug¬
menting path, shown at the top of Figure 10.8. The three horizontal edges, in
order from left to right, now have residual capacities 1, 0, and 0. Suppose
inductively that the horizontal residual capacities are 0 fc ~ 1 , 0, and 0 k for some
non-negative integer k.

1. Augment along path B, adding f> k to the flow; the residual capacities are
now 4> k+1 , 4> k , and 0.

2. Augment along path C, adding 0 fc to the flow; the residual capacities are
now 0 fc+1 , 0, and 0 fc .

3. Augment along path B, adding 0 k +1 to the flow; the residual capacities
are now 0, <fi k+1 , and cp k+2 .

4. Augment along path A, adding <f k+1 to the flow; the residual capacities
are now 0 ,c+1 , 0, and 0 fc+2 .

It follows by induction that after 4n + 1 augmentation steps, the horizontal
edges have residual capacities 0 2n_2 ,0, 0 2n_1 . As the number of augmentations
grows to infinity, the value of the flow converges to

1 +

2l>' =

1 +

;=i

1-0

= 4 + a/5 < 7,

even though the maximum flow value is clearly 2X + 1 >■ 7.

Practically-minded readers might wonder why anyone should care about
irrational capacities; after all, computers can’t represent anything but (small)
integers or (small dyadic) rationals exactly. Good question! The mathematician’s
answer is that the restriction to integer capacities is literally artificial; it’s an
artifact of digital computational hardware (or perhaps the otherwise irrelevant
laws of physics), not an inherent feature of the abstract computational problem.
But a more practical reason is that the behavior of the algorithm with irrational
inputs tells us something about its worst-case behavior in practice with floating¬
point capacities—terrible! Even with very reasonable capacities, a careless
implementation of Ford-Fulkerson could enter an infinite loop, simply because
of round-off error, without ever coming close to the correct answer.

10.5 Combining and Decomposing Flows

Flows are normally defined as functions on the edges of a graph satisfying certain
constraints at the vertices. However, flows have a second characterization that
is more natural and useful in certain contexts.

Consider an arbitrary graph G with source vertex s and target vertex t. Fix
any two (s, t)-flows / and g and any two real numbers a and f>, and consider

334

10.5. Combining and Decomposing Flows

the function h: E —> K defined by setting

h(u-yv') := a ■ f(u->v) + fi ■ g(u-»v)

for every edge u-*v\ we can write this definition more simply as h = af + fig.
Straightforward definition-chasing implies that h is also an (s, t)-flow with value
\h\ = a\f\ + fi\g\. More generally, any linear combination of (s, t)-flows is also
an (s, t)-flow.

It turns out that any (s, t)-flow can be written as a weighted sum of flows
with a very special structure. For any directed path P from s to t, we define a
corresponding path flow as follows:

P(u->v)

1 ifu^veP,
< —1 if v^u € P,
0 otherwise.

Straightforward definition-chasing implies that the function P: E —> R is indeed
an (s, t)-flow with value 1. I am deliberately overloading the variable P to mean
both the path (a sequence of vertices and directed edges) and the unit flow
along that path.

Similarly, for any directed cycle C, we define a corresponding cycle flow by
setting

C(u^v)

1 if u—>v e C,

< —1 if v—m € C,
0 otherwise.

v.

Again, it is easy to verify that C: E —» R is an (s, t)-flow with value zero.

Our earlier argument implies that any linear combination of path flows and
cycle flows is another flow; this weighted sum is called a flow decomposition.
Moreover, every non-negative flow has a flow decomposition with the following
special structure.

Flow Decomposition Theorem. Every non-negative (s, t)-flow f can be writ¬
ten as a positive linear combination of directed (s, t)-paths and directed cycles.
Moreover, a directed edge u->v appears in at least one of these paths or cycles
if and only if f (u->v) > 0, and the total number of paths and cycles is at most
the number of edges in the network.

Proof: We prove the theorem by induction on the number of edges carrying
non-zero flow, intuitively by running the Ford-Fulkerson algorithm backward.
As long as at least one edge in the graph carries positive flow, we can find either
an (s, t)-path or a directed cycle that carries flow. Subtracting as much flow

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io. Maximum Flows & Minimum Cuts

Figure lo.g. Decomposing a circulation into weighted directed cycles.

as possible from that path or cycle empties at least one edge, so the Recursion
Fairy can give us the rest of the decomposition.

To formalize this argument, we first consider the special case of circulations;
these are flows with value 0, where flow is conserved at every vertex. Fix an
arbitrary circulation / in an arbitrary flow network, and let #/ denote the
number of edges u->v such that f{u->v ) >0.1 claim that / can be decomposed
into a positive linear combination of at most max{0, #/ — 1} cycles. There are
three cases to consider:

• If #/ = 0, then / is vacuously a linear combination of zero cycles.

• Suppose f{u->v ) > 0 for a single directed cycle of edges u->v. Then #/ > 2,
and / is trivially a linear combination of one cycle.

• Otherwise, pick an arbitrary edge u->v with /(u->v) > 0. Consider an arbi¬
trary walk v Q —>v 1 ■ ■ ■ with v 0 = u and v 1 = v, such that/(v i _ 1 ^v i ) > 0

for every index i. The conservation constraint implies that every vertex with
incoming flow also has outgoing flow, so we can make this walk arbitrarily
long; in particular, the walk must eventually visit some vertex more than
once. Let j < k be the smallest indices such that v } = v k . The subwalk
Vj->Vj_ t -> - >v k is a simple directed cycle C.

Define F := min eeC /(e), and consider the function f':=f-F-C, or
more verbosely,

f/(u-»v) — F ifiz^veC,

fXu^v) :=

I /(u->v) otherwise.

336

Straightforward definition chasing shows that f is a feasible circulation
in G with value 0. There is at least one edge e e C such that /(e) = F, and

10.5. Combining and Decomposing Flows

therefore /'(e) = 0, which implies #/' < #/ — 1. Since fewer edges carry
flow in f than in /, the induction hypothesis implies that f has a valid
decomposition into #f — 1 < #/ — 2 cycles. Adding F units of flow around
cycle C gives us a flow decomposition for /; more succinctly: / = f' + F ■ C.

Now let / be an arbitrary (s, t)-flow in an arbitrary flow network. Add an
edge t->s to the network, and define a circulation f by setting /'(t->s) = |/|
and f'(u->v ) = f{u->v ) for every original edge u-»v. The previous argument
implies that the circulation /' is a positive linear combination of at most #f' — 1
directed cycles. Deleting the edge t->s gives us a decomposition of the original
flow / into at most #/' — 1 < #/ paths and cycles. Specifically, cycles in f' that
include t->s become (s, t)-paths in /, and cycles in f that do not include t-«
remain cycles in /. □

The proof of the Flow Decomposition Theorem implies stronger results in
two interesting special cases.

• Any circulation can be decomposed into a weighted sum of cycles; no paths
are necessary.

• Any acyclic (s, t)-flowcanbe decomposed into a weighted sum of (s, t)-paths;
no cycles are necessary.

Moreover, by canceling flow cycles until no more remain, we can transform any
flow into an acyclic flow with the same value. In particular, every flow network
supports a maximum (s, t)-flow that is acyclic.

The proof also immediately translates directly into an algorithm, similar
to Ford-Fulkerson, to decompose any (s, t)-flow into paths and cycles. The
algorithm repeatedly seeks either a directed (s, t)-path or a directed cycle in the
remaining flow, and then subtracts as much flow as possible along that path or
cycle, until the flow is empty. We can find a flow path or cycle in 0(V ) time as
follows:

• If any edge leaving s has positive flow, follow an arbitrary walk from s in the
flow graph until it either reaches t (giving us a flow path) or reaches some
vertex for the second time (giving us a flow cycle).

• If no edge leaving s has positive flow, find any other vertex v with positive
outflow, and follow an arbitrary walk from v in the flow graph until it reaches
some vertex for the second time (giving us a flow cycle).

In both cases, the conservation constraint implies that this algorithm will never
get stuck. Because each iteration removes at least one edge from the flow graph,
the entire decomposition algorithm runs in 0(VE ) time.

Flow decompositions provide a natural lower bound on the running time of
any maximum-flow algorithm that builds the flow one path or cycle at a time.
Every flow can be decomposed into at most E paths and cycles, each of which uses

337

io. Maximum Flows & Minimum Cuts

at most V edges, so the overall complexity of the flow decomposition is 0(VE).
Moreover, it is easy to construct flows for which every flow decomposition
has complexity El(VE). Thus, any maximum-flow algorithm that explicitly
constructs a flow one path or cycle at a time—in particular, any implementation
of Ford and Fulkerson’s augmenting path algorithm—must take D(VE) time in
the worst case.

10.6 Edmonds and Karp's Algorithms

Ford and Fulkerson’s algorithm does not specify which path in the residual
graph to augment; the poor worst-case behavior of the algorithm can be blamed
on poor choices for the augmenting path. In the early 1970 s, Jack Edmonds and
Richard Karp analyzed two natural rules for choosing augmenting paths, both
of which led to more efficient algorithms.

Fattest Augmenting Paths

Edmonds and Karp’s first rule is essentially a greedy algorithm:

Choose the augmenting path with largest bottleneck value.

It’s not hard to show that the maximum-bottleneck (s, t)-path in a directed graph
can be computed in 0(E log!/) time using a “best-first” traversal, similar to
Jarnik’s minimum-spanning-tree algorithm or Dijkstra’s shortest-path algorithm.
The algorithm grows a directed tree T, rooted at s, one vertex at a time, by
repeatedly adding the highest-capacity edge leaving T to T, until T contains
a path from s to t. Alternately, one could emulate Kruskal’s algorithm—insert
edges one at a time in decreasing capacity order until there is a path from s
to t —although this approach is less efficient, at least when the graph is directed.

To complete the running-time analysis of the flow algorithm, we need an
upper bound on the number of iterations before the algorithm halts. In fact,
for arbitrary real capacities, the algorithm may never halt; see Exercise 18 . For
integer capacities, however, we can bound the number of iterations as a function
of the maximum flow value |/*|, as follows.

Let / be any flow in G, and let f' be the maximum flow in the current
residual graph Gj . (At the beginning of the algorithm, Gf = G and f = /*.)
We have already proved that f can be decomposed into at most E paths and
cycles. A simple averaging argument implies that at least one of the paths in this
decomposition must carry at least \f'\/E units of flow. It follows immediately
that the fattest (s, t)-path in Gf carries at least | f'\/E units of flow.

Thus, augmenting / along the maximum-bottleneck path in Gf multiplies
the value of the remaining maximum flow in Gf by a factor of at most 1 — 1 /E.

338

10 . 6 . Edmonds and Karp's Algorithms

In other words, the residual maximum flow value decays exponentially with the
number of iterations. After E ■ ln|/*| iterations, the maximum flow value in Gf
is at most

|/*| • (1 - l/E) JMn| - ri < |/*| e~ ln{n = 1.

(That’s Euler’s constant e, not the edge e. Sorry.) In particular, after E ■ ln|/*|
iterations, the residual maximum flow value is less than 1. If all capacities are
integers, the residual maximum flow value is also an integer, so it must be 0; in
other words, / is a maximum flow!

We conclude that for graphs with integer capacities, the Edmonds-Karp
“fattest path” algorithm runs in 0(E 2 logE log|/*|) time. Unlike the worst-case
running time of raw Ford-Fulkerson, this time bound is actually a polynomial
function of the input size.

Just like the original Ford-Fulkerson algorithm, the "fattest path" algorithm
can get stuck in an infinite loop in networks with arbitrary real capacities.
However, our analysis implies that even if the algorithm never halts, it maintains
a flow / that approaches a maximum flow in the limit.

Shortest Augmenting Paths

The second Edmonds-Karp rule was actually proposed by Ford and Fulkerson in
their original maximum-flow paper; a variant of this rule was independently
considered by the Russian mathematician Yefim Dinitz 3 around the same time
as Edmonds and Karp.

Choose the augmenting path with the smallest number of edges.

The shortest augmenting path can be found in 0(F) time by running breadth-first
search in the residual graph. Surprisingly, the resulting algorithm halts after a
polynomial number of iterations, independent of the actual edge capacities!

The proof of this polynomial upper bound relies on two observations
about the evolution of the residual graph. Let /, be the current flow after i
augmentation steps, let G ; be the corresponding residual graph. In particular,
/ 0 is zero everywhere and G 0 = G. For each vertex v, let level^v) denote the
unweighted shortest-path distance from s to v in G 1; or equivalently, the level of
v in a breadth-first search tree of G ; rooted ats. In particular, if there is no path
from s to v in G ; , then levelfv) = oo (because min0 = oo).

Our first observation is that the level of a vertex can only increase over time.

Lemma 10 . 2 . levelfv) > level i _ 1 (v) for all vertices v and all integers i > 0.

3 Dinitz actually discovered an even faster maximum-flow algorithm that runs in 0{V 2 E )
time, while a student in an algorithms class taught by Georgy Adelson-Velsky (the “AV” in AVL
trees), in response to an in-class exercise.

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io. Maximum Flows & Minimum Cuts

Proof: Fix an arbitrary positive integer i > 0 and an arbitrary vertex v. We
prove the claim by induction on level;(v) (and not on the integer i). As an
inductive hypothesis, assume for every vertex u such that level^u) < level ^v),
that level; (u) > Zevei i _ 1 (u). There are three cases to consider.

• If v = s, we immediately have level; (s) = levels (s) = 0.

• If there is no path from s to v in G ; , then level;(v) = oo > Zevel i _ 1 (v).

• Otherwise, let s-> ->u->v be any unweighted shortest path from s to v in

the graph G;. Because this is a shortest path, we have level;(v) = level ; (u) +1,
so the inductive hypothesis implies level;(iz) > level;.; (u). To complete the
proof, we need to show that level i _ 1 (u) > level;_ 1 (v)— 1. We have two
subcases to consider.

- If u—>v is an edge in G;_ 1; then level i _ 1 (v) < level i _ 1 (u) + 1, because the
levels are defined by breadth-first traversal.

- On the other hand, if u^v is not an edge in G;_ l5 then its reversal
v—>tz must be an edge in the 1 th augmenting path, which by definition
is the shortest path from s to t in G;_;. It follows that level^fv) =
level i _ 1 (u) — 1 < levelj.^tz) + 1 .

In both subcases, we conclude that level;(v) = level;(u)+1 > leveli_i(u ) + 1 >
levelj.jtv). □

Whenever we augment the flow, the bottleneck edge in the augmenting
path disappears from the residual graph, and some edges in the reversal of the
augmenting path may (re-) appear. Our second observation is that an edge
cannot appear or disappear too many times.

Lemma 10 . 3 . During the execution of the Edmonds-Karp shortest-augmenting-
path algorithm, each edge u^v disappears from the residual graph Gj at most
V /2 times.

Proof: Suppose u->v is in two residual graphs G, : and G (+1 , but not in any of
the intermediate residual graphs G i+1 ,..., Gj, for some 1 < j. Then u^v must
be in the 1 th augmenting path, so level ; (v) = level ; (u) + 1 , and v^iz must be
on the jth augmenting path, so level ; (v) = level } (u) — 1. The previous lemma
implies that

levelj{u ) = level J (v) + 1 > level ; (v) + 1 = level ! (u) + 2 .

In other words, between the disappearance and reappearance of iz^v, the
distance from s to u increased by at least 2. Because every level is either less
than V or infinite, the number of disappearances is at most V/2. □

340

107- Further Progress

Now we can derive an upper bound on the number of iterations. Because each
edge disappears at most V/2 times, there are at most EV / 2 edge disappearances
overall. But at least one edge disappears on each iteration, so the algorithm
must halt after at most EV /2 iterations. Finally, each iteration requires 0(E)
time, so the overall algorithm runs in 0(VE 2 ) time.

10.7 Further Progress

This is nowhere near the end of the story for maximum-flow algorithms. Decades
of further research have led to several faster algorithms, some of which are
summarized in Figure 10 . 10. 4 All the listed algorithms listed compute a
maximum flow in several iterations. Most of these algorithms have two variants:
a simpler version that performs each iteration by brute force, and a faster
variant that uses sophisticated data structures to maintain a spanning tree of
the flow network, so that each iteration can be performed (and the spanning
tree updated) in logarithmic time. There is no reason to believe that the best
algorithms known so far are optimal; indeed, maximum flows are still a very
active area of research.

Technique

Direct

With dynamic trees

Source(s)

Blocking flow

o{v 2 e )

0(VE logV)

[Dinitz; Karzanov; Even and Itai;
Sleator and Tarjan]

Network simplex

0(V 2 E )

0(VE logV)

[Dantzig; Goldfarb and Plao;
Goldberg, Grigoriadis, and Tarjan]

Push-relabel (generic)

0(V 2 E )

-

[Goldberg and Tarjan]

Push-relabel (FIFO)

0 (V 3 )

0 (VElog(V 2 /E))

[Goldberg and Tarjan]

Push-relabel (highest label)

o(v 2 Ve )

-

[Cheriyan and Maheshwari; Tungel]

Push-relabel-add games

—

0(VE log B /(i, logV ) V )

[Cheriyan and Plagerup;

King, Rao, and Tarjan]

Pseudoflow

0(V 2 E )

0 (VE logV)

[Plochbaum]

Pseudoflow (highest label)

0 (V 3 )

0 (V£log(V 2 /E))

[Plochbaum and Orlin]

Incremental BFS

0(V 2 E )

0 (VE log(V 2 /E))

[Goldberg, Field, Kaplan, Tarjan,
and Werneck]

Compact networks

-

0 (V£)

[Orlin]

Figure 10.10. Several purely combinatorial maximum-flow algorithms and their running times.

The fastest known (purely combinatorial) maximum-flow algorithm, an¬
nounced by James Orlin in 2012 , runs in 0(VE) time, exactly matching the
worst-case complexity of a flow decomposition. The details of Orlin’s algorithm

4 To keep this table short, I have deliberately omitted algorithms whose running time depends
on edge capacities or the maximum flow value. Even with this restriction, the list is embarrassingly
incomplete!

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io. Maximum Flows & Minimum Cuts

are far beyond the scope of this book; in addition to his own new techniques,
Orlin uses several older algorithms and data structures as black boxes, most of
which are themselves quite complicated. In particular, Orlin’s algorithm does
not construct an explicit flow decomposition; in fact, for graphs with only 0(V )
edges, an extension of his algorithm actually runs in only 0(V 2 /log V) time!
Nevertheless, for purposes of analyzing algorithms that use maximum flows,
this is the time bound you should cite. So write the following sentence on your
cheat sheets and cite it in your homeworks:

Maximum flows can be computed in 0(VE) time.

Finally, faster maximum-flow algorithms are known for unit-capacity net¬
works, where every edge has capacity 1. In 1973 , Alexander Karzanov proved
that Dinitz’s blocking-flow algorithm—the first algorithm listed in the table
above—runs in 0(mm{V 2 ^ ,E l ^ 2 }E) time in this setting. (This time bound
appears to break the Gt(VE) flow decomposition barrier, but in fact Karzanov’s
analysis implies that any flow in a unit-capacity network can be decomposed
into paths with total complexity 0(mm{V 2 ^,E 1 ^ 2 } E).) This was the fastest
algorithm known in this setting for more than 40 years. Karzanov’s record
was finally broken in 2014 , when Aleksander Mqdry announced a truly remark¬
able algorithm that computes maximum flows in unit-capacity networks in
0(£ 10 ' 7 polylog £) time. Again, the details of Mqdry’s algorithm are far beyond
the scope of this book, or indeed the expertise of its author.

Exercises

o. Suppose you are given a directed graph G = (V,E), two vertices s and t,
a capacity function c: E —* R + , and a second function / : E —» R. Describe
an algorithm to determine whether / is a maximum (s, t)-flow in G.

1 . Let / and f be two feasible (s, t)-flows in a flow network G, such that
\f'\ > 1/1- Prove that there is a feasible (s, t)-flow with value \f'\ — \f | in the
residual network Gf.

2 . Let u—>v be an arbitrary edge in an arbitrary flow network G. Prove that if
there is a minimum (s, t)-cut (S, T ) such that u € S and v e T, then there is
no minimum cut ( S', T') such that u&T' and v € S'.

3. Let (S, T ) and (S', T') be minimum (s, t)-cuts in some flow network G. Prove
that (S nS'jU T') and (S U S', T n T') are also minimum (s, t)-cuts in G.

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Exercises

4 . Let G be a flow network that contains an opposing pair of edges tz->v and
v^u, both with positive capacity. Let G' be the flow network obtained from G
by decreasing the capacities of both of these edges by min{c(u- 4 v), c(v-uz)}.
In other words:

• If c(u—>v) > c(v^u), change the capacity of u-*v to c(u->v) — c(v^u)
and delete v^u.

• If c(u—>v) < c(v^u), change the capacity of v-m to c(v^u) — c(u->v )
and delete u->v.

• Finally, if c(u^v) = c(v->u), delete both u->v and v-ni.

Figure 10.11. Enforcing the one-direction assumption.

(a) Prove that every maximum (s, t)-flow in G' is also a maximum flow in G.
(Thus, by reducing every opposing pair of edges in G, we obtain a new
flow network without opposing edges, but with the same maximum flow
value as G.)

(b) Prove that every minimum (s, t)-cut in G is also a minimum (s, t)-cut
in G' and vice versa.

(c) Prove that there is at least one maximum (s, t)-flow in G that is not a
maximum (s, t)-flow in G'.

5 . (a) Describe an efficient algorithm to determine whether a given flow

network contains a unique maximum (s, t)-flow.

(b) Describe an efficient algorithm to determine whether a given flow
network contains a unique minimum (s, t)-cut.

(c) Describe a flow network that contains a unique maximum (s, t)-flow but
does not contain a unique minimum (s, t)-cut.

(d) Describe a flow network that contains a unique minimum (s, t)-cut but
does not contain a unique maximum (s, t)-flow.

6 . An (s, t)-flow in a network G is acyclic if there are no directed cycles where
every edge has a positive flow value; that is, the subgraph of edges with
positive flow value is a dag.

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io. Maximum Flows & Minimum Cuts

(a) Describe and analyze an algorithm to compute an acyclic maximum
(s, t)-flow in a given flow network. Your algorithm should have the same
asymptotic running time as Ford-Fulkerson.

(b) Describe and analyze an algorithm to determine whether every maximum
(s, t)-flow in a given flow network is acyclic.

7 . Let G = ([V,E ) be a flow network in which every edge has capacity 1 and
the shortest-path distance from s to t is at least d.

(a) Prove that the value of the maximum (s, t)-flow is at most E/d.

(b) Now suppose that G is simple, meaning that for all vertices u and v, there
is at most one edge from u to v. (Flow networks can have parallel edges.)
Prove that the value of the maximum (s, t)-flow is at most 0(V 2 /d 2 ).
[Hint: How many nodes are in the average level of a BFS tree rooted
ats?]

8 . Suppose we are given a flow network G = (V, E) in which every edge has
capacity 1, together with an integer k. Describe and analyze an algorithm
to identify k edges in G such that after deleting those k edges, the value of
the maximum (s, t)-flow in the remaining graph is as small as possible.

9 . The analysis in our proof of the Flow Decomposition Theorem can be
tightened. Let G = (V,E) be an arbitrary flow network, and let / be an
arbitrary (s, t)-flow in G.

(a) Prove that if \f \ = 0, then / is the weighted sum of at most E — V + 1
directed cycles, where /(e) > 0 for every edge e in each of these cycles.

(b) Prove that if \f | > 0, then / is the weighted sum of at most E — V + 2
directed paths and directed cycles, where /(e) > 0 for every edge e in
each of these paths and cycles.

(c) Prove that both of the previous upper bounds are tight—some circulations
cannot be decomposed into less than E — V + 1 cycles, and some flows
cannot be decomposed into less than E — V + 2 paths and cycles. [Hint:
This is easy.]

* 10 . Our observation that any linear combination of (s, t)-flows is itself an (s, t)-
flow implies that the set of all (not necessarily feasible) (s, t)-flows in any
graph actually define a real vector space, which we can call the/Zow space of
the graph.

(a) Prove that the flow space of any connected graph G = (V, E ) has dimen¬
sion E — V + 2.

344

Exercises

(b) Let T be any spanning tree of G. Prove that the following collection of
paths and cycles define a basis for the flow space:

• The unique path in T from s to t;

• The unique cycle in T U {e}, for every edge e ^ T.

(c) Let T be any spanning tree of G, and let F be the forest obtained by
deleting any single edge in T. Prove that the following collection of
paths and cycles define a basis for the flow space:

• The unique path in F U {e} from s to t, for every edge e ^ F that has
one endpoint in each component of F ;

• The unique cycle in F U {e}, for every edge e ^ F with both endpoints
in the same component of F.

(d) Prove or disprove the following claim: Every connected flow network
has a flow basis that consists entirely of simple paths from s to t.

n. Cuts are sometimes defined as subsets of the edges of the graph, instead of
as partitions of its vertices. In this problem, you will prove that these two
definitions are almost equivalent.

We say that a subset X of (directed) edges separates s and t if every
directed path from s to t contains at least one (directed) edge inX. For any
subset S of vertices, let 5S denote the set of directed edges leaving S; that is,
5S := {u^v | iz € S, v ^ S}.

(a) Prove that if (S, T) is an (s, t)-cut, then 5S separates s and t.

(b) LetX be an arbitrary subset of edges that separates s and t. Prove that
there is an (s, t)-cut (S, T ) such that 5S c X.

(c) LetX be a minimal subset of edges that separates s and t. (Such a set
of edges is sometimes called a bond.) Prove that there is an (s, t)-cut
(S, T) such that 5S = X.

12 . Suppose instead of capacities, we consider networks where each edge u^v
has a non-negative demand d(u->v). Now an (s, t)-flow/ is feasible if and
only if f(u->v) > d(u^>v) for every edge u->v. (Feasible flow values can now
be arbitrarily large.) A natural problem in this setting is to find a feasible
(s, t)-flow of minimum value.

(a) Describe an efficient algorithm to compute a feasible (s, t)-flow, given
the graph, the demand function, and the vertices s and t as input. [Hint:
Find a flow that is non-zero everywhere, and then scale it up to make it
feasible.]

(b) Suppose you have access to a subroutine MaxFlow that computes
maximum flows in networks with edge capacities. Describe an efficient

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io. Maximum Flows & Minimum Cuts

algorithm to compute a minimum flow in a given network with edge
demands; your algorithm should call MaxFlow exactly once.

(c) State and prove an analogue of the max-flow min-cut theorem for this
setting. (Do minimum flows correspond to maximum cuts?)

13. For any flow network G and any vertices u and v, let bottleneck G {u, v) denote
the maximum, over all paths n in G from u to v, of the minimum-capacity
edge along n.

(a) Describe and analyze an algorithm to compute bottleneck^, t) in
0(E logV) time. This is the amount of flow that the Edmonds-Karp
fattest-augmenting-paths algorithm pushes in the first iteration.

(b) Now suppose the flow network G is undirected; equivalently, suppose
c(tz^v) = c(v-m) for every pair of vertices u and v. Describe and analyze
an algorithm to compute bottleneck G (s, t) in 0(E) time. [Hint: Find the
median edge capacity.] Why doesn’t this speedup work for directed
graphs?

v (c) Again, suppose the flow network G is undirected. Describe and an¬
alyze an algorithm to construct a spanning tree T of G such that
bottleneck T {u, v) = bottleneck G {u, v) for all vertices u and v. (Edges in T
inherit their capacities from G.) For full credit, your algorithm should
run in 0(E) time.

14. Suppose you are given a flow network G with integer edge capacities and
an integer maximum flow f * in G. Describe algorithms for the following
operations:

(a) iNCREMENT(e): Increase the capacity of edge e by 1 and update the
maximum flow.

(b) Decrement^ ): Decrease the capacity of edge e by 1 and update the
maximum flow.

Both algorithms should modify/* so that it is still a maximum flow, more
quickly than recomputing a maximum flow from scratch.

15. Let G be a network with integer edge capacities. An edge in G is upper¬
binding if increasing its capacity by 1 also increases the value of the maximum
flow in G. Similarly, an edge is lower-binding if decreasing its capacity by 1
also decreases the value of the maximum flow in G.

(a) Does every network G have at least one upper-binding edge? Prove your
answer is correct.

(b) Does every network G have at least one lower-binding edge? Prove your
answer is correct.

346

Exercises

(c) Describe an algorithm to find all upper-binding edges in G, given both G
and a maximum flow in G as input, in 0(E) time.

(d) Describe an algorithm to find all lower-binding edges in G, given both G
and a maximum flow in G as input, in O(EV) time.

r6. A given flow network G may have more than one minimum (s, t)-cut. Let’s
define the best minimum (s, t)-cut to be any minimum cut ( S, T ) with the
smallest number of edges crossing from S to T.

(a) Describe an efficient algorithm to find the best minimum (s, t)-cut when
the capacities are integers.

(b) Describe an efficient algorithm to find the best minimum (s, t)-cut for
arbitrary edge capacities.

(c) Describe an efficient algorithm to determine whether a given flow
network contains a unique best minimum (s, t)-cut.

17. A new assistant professor, teaching maximum flows for the first time, suggests
the following greedy modification to the generic Ford-Fulkerson augmenting
path algorithm. Instead of maintaining a residual graph, just 5 reduce the
capacity of edges along the augmenting path! In particular, whenever we
saturate an edge, just remove it from the graph. Who needs all that residual
graph nonsense?

GreedyFlow(G,c,s, t):
for every edge e in G
/(e)«- 0

while there is a path from s to t

n <— an arbitrary path from s to t
F <— minimum capacity of any edge in n
for every edge e in n
/(e)«-/(e) + F
if c(e) = F

remove e from G

else

c(e) <— c(e) — F

return /

(a) Show that GreedyFlow does not always compute a maximum flow.

(b) Show that GreedyFlow is not even guaranteed to compute a good
approximation to the maximum flow. That is, for any constant a > 1 ,
there is a flow network G such that the value of the maximum flow is

5 More often than not, the adverb just is subconscious shorthand for “I’m too lazy to figure
out the details, so this is almost certainly wrong, but you should believe me anyway.” See also
merely, simply, clearly, and obviously.

347

io. Maximum Flows & Minimum Cuts

more than a times the value of the flow computed by GreedyFlow.
[Hint: Assume that GreedyFlow chooses the worst possible path n at
each iteration.]

18. In 1980 Maurice Queyranne published an example of a flow network, shown
below, where Edmonds and Karp’s “fattest path” heuristic does not halt. As
in Zwick’s bad example for the original Ford-Fulkerson algorithm, (p denotes
the inverse golden ratio (\/ 5—1 )/ 2 . The three vertical edges play essentially
the same role as the horizontal edges in Zwick’s example.

Figure 10.12. Queyranne's network, and a sequence of "fattest path" augmentations.

(a) Show that the following infinite sequence of path augmentations is a
valid execution of the Edmonds-Karp “fattest path” algorithm. (See the
bottom of Figure 10.12.)

QueyranneFatPaths :
for i <— 1 to 00

push cp 3l ~ 2 units of flow along s^a^f ^g^b^h^c^d^t
push (p 3l ~ 3 units of flow along s->/ ^a^b^g^h^c^t
push <p 31 units of flow along s->e^/ -^a^g^b^c^h-^-t

(b) Describe a sequence of 0 ( 1 ) path augmentations that yields a maximum
flow in Queyranne’s network.

19. An (s, t)-series-parallel graph is a directed acyclic graph with two distin¬
guished vertices s and t and with one of the following structures:

• Base case: A single directed edge from s to t.

• Series: The union of an (s, u)-series-parallel graph and a (u, ^-series-
parallel graph that share a common vertex u but no other vertices or
edges.

• Parallel: The union of two smaller (s, t)-series-parallel graphs with
the same source s and target t, but with no other vertices or edges in
common.

348

Exercises

Every ( 5 , t)-series-parallel graph G can be represented by a decomposition
tree, which is a binary tree with three types of nodes: leaves (which
corresponding to edges in G), series nodes (which correspond to vertices
other than s and t), and parallel nodes. The same series-parallel graph
could be represented by many different decomposition trees.

Figure 10.13. A series-parallel graph and a corresponding decomposition tree. Squares in the decom¬
position tree are leaves; diamonds are parallel nodes.

(a) Suppose you are given a directed graph G with two special vertices s and t.
Describe and analyze an algorithm that either builds a decomposition
tree for G or correctly reports that G is not (s, t)-series-parallel. [Hint:
Build the tree from the bottom up.]

(b) Describe and analyze an algorithm to compute a maximum (s, t)-flow in
a given (s, t)-series-parallel flow network with arbitrary edge capacities.
[Hint: In light of part (a), you can assume that you are actually given
the decomposition tree. First compute the maximum-flow value, then
compute an actual maximum flow.]

20 . We can speed up the Edmonds-Karp “fattest path” heuristic, at least for
networks with small integer capacities, by relaxing our requirements for
the next augmenting path. Instead of finding the augmenting path with
maximum bottleneck capacity, we find a path whose bottleneck capacity is
at least half of maximum, using the following capacity scaling algorithm.

The algorithm maintains a bottleneck threshold A; initially, A is the
maximum capacity among all edges in the graph. In each phase, the
algorithm augments along paths from s to t in which every edge has residual
capacity at least A. When there is no such path, the phase ends, we set
A <— [A/2J, and the next phase begins. The algorithm ends when A = 0.

(a) How many phases will the algorithm execute in the worst case, if the
edge capacities are integers?

349

io. Maximum Flows & Minimum Cuts

(b) Let / be the flow at the end of a phase for a particular value of A. Prove
that the capacity of a minimum cut in the residual graph Gf is at most
E ■ A.

(c) Prove that in each phase of the scaling algorithm, there are at most 2 E
augmentations.

(d) What is the overall running time of this capacity scaling algorithm,
assuming all the edge capacities are integers less than U ?

350

For a long time it puzzled me how something so expensive, so leading edge, could be so
useless, and then it occurred to me that a computer is a stupid machine with the ability
to do incredibly smart things, while computer programmers are smart people with the
ability to do incredibly stupid things. They are, in short, a perfect match.

- Bill Bryson, Notes from a Big Country (1999)

Shortly after the "iron curtain" fell in t ggo, an American and a Russian, who had both
worked on the development of weapons, met. The American asked: "When you
developed the Bomb, how were you able to perform such an enormous amount of
computing with your weak computers?" The Russian responded: "We used better
algorithms."

— Yefim Dinitz, in "Dinitz' Algorithm: The Original Version
and Even's Version” (2006)

So i'm the bad guy now I hear, because I don't go with the flow.

Don't ever go with the flow, be the flow.

— Jay-Z, "Stream of Consciousness", May 16, 2015

Applications of Flows and Cuts

11.1 Edge-Disjoint Paths

One of the easiest applications of maximum flows is computing the maximum
number of edge-disjoint paths between two specified vertices s and t in a
directed graph G using maximum flows. A set of paths in G is edge-disjoint if
each edge in G appears in at most one of the paths; several edge-disjoint paths
may pass through the same vertex, however.

If we give each edge capacity 1, then the maximum flow from s to t pushes
either 0 or 1 units of flow along each edge. The flow-decomposition theorem
implies that the subgraph S of saturated edges is the union of several edge-
disjoint paths and cycles. Moreover, the number of paths in this decomposition
is exactly equal to the value of the flow. Extracting the actual paths from S is
easy—follow any directed path in S from s to t, remove that path from S, and
recurse.

Conversely, we can transform any collection of k edge-disjoint paths into a

35i

ii. Applications of Flows and Cuts

flow by pushing one unit of flow along each path from s to t; the value of the
resulting flow is exactly k. It follows that any maximum flow algorithm actually
computes the largest possible set of edge-disjoint paths.

If we use Orlin’s algorithm to compute maximum flows, we can compute
edge-disjoint paths in 0(VE ) time, but Orlin’s algorithm is overkill for this
simple application. The cut ({s}, V \ {s}) has capacity at most V — 1, so the
maximum flow has value at most V — l. Thus, Ford and Fulkerson’s original
augmenting path algorithm already runs in 0([/*| E) = 0(VE) time.

The same algorithm can also be used to find edge-disjoint paths in undirected
graphs. First, replace every undirected edge uv in G with a pair of directed
edges u->v and v->u, each with unit capacity, and call the resulting directed
graph G'. Next, compute a maximum (s, t)-flow /* in G' using Ford-Fulkerson.
For any edge uv in G, if /* saturates both directed edges u-»v and v^u in G',
we can remove both edges from the flow without changing its value. (More
generally, we can find an acyclic maximum flow in G' by canceling all cycles
in /*, not only cycles of length 2.) Thus, without loss of generality, f* assigns a
unique direction to each saturated edge. Finally, we can extract the edge-disjoint
paths by searching the subgraph of directed edges saturated by /*.

11.2 Vertex Capacities and Vertex-Disjoint Paths

Now suppose the vertices of the input graph G have capacities, not just the
edges. In addition to our other constraints, for each vertex v other than s and t,
we require the total flow into v (and therefore the total flow out of v) to be at
most some non-negative value c(v):

Can we still compute maximum flows with these new vertex constraints?

In 1962 , Ford and Fulkerson proposed the following reduction to a flow
network G with only edge capacities. Replace every vertex v with two vertices
v in and v out , connected by an edge v in ->v out with capacity c(v), and then replace
every directed edge u^v with the edge u 0 Ut ^v in (keeping the same capacity).
Routine definition-chasing implies that every feasible (.s 0llt , t in )-flow in G is
equivalent to a feasible (s, t)-flow with the same value in the original graph G,
and vice versa. In particular, every maximum flow in G is equivalent to a
maximum flow in G. The reduction from G to G takes 0(F) time, after which
we can compute the maximum flow in G using Orlin’s algorithm. Altogether,
computing the maximum flow in G requires 0(VE ) time.

It is now easy to compute the maximum number of vertex-disjoint paths
from s to t in 0(VE ) time: Assign capacity 1 to every vertex and compute a
maximum flow!

352

11.3. Bipartite Matching

Figure 11.1 . Reducing vertex-disjoint paths in G to edge-disjoint paths in G.

11.3 Bipartite Matching

Another natural application of maximum flows is finding maximum matchings
in bipartite graphs. A matching is a subgraph in which every vertex has degree
at most one, or equivalently, a collection of edges such that no two share a
vertex. The problem is to find a matching with the maximum number of edges.

For example, suppose we have a set of doctors who are looking for jobs, and
a set of hospitals who are looking for doctors. Each doctor lists all hospitals
where they are willing to work, and each hospital lists all doctors they are
willing to hire. Our task is to find the largest subset of doctor-hospital hires that
everyone is willing to accept . 1 This problem is equivalent to finding a maximum
matching in a bipartite graph whose vertices are the doctors and hospitals, and
there is an edge between a doctor and a hospital if and only if each find the
other acceptable.

We can solve this problem by reducing it to a maximum flow problem, as
follows. Let G be the given bipartite graph with vertex set LUR, such that every
edge joins a vertex in L to a vertex in R. We create a new directed graph G'
by ( 1 ) orienting each edge from L to R, ( 2 ) adding two new vertices s and t,
( 3 ) adding edges from s to every vertex in U, and ( 4 ) adding edges from each
vertex in R to t. Finally, we assign every edge in G' a capacity of 1.

Any matching M in G can be transformed into a flow f M in G' as follows:
For each edge uw in M, push one unit of flow along the path s^u^w^t. These
paths are disjoint except at s and t, so the resulting flow satisfies the capacity
constraints. Moreover, the value of the resulting flow is equal to the number of
edges in M.

Conversely, consider any (s, t)-flow / in G', computed using the Ford-
Fulkerson augmenting path algorithm. Because the edge capacities are integers,
the Ford-Fulkerson algorithm assigns an integer flow to every edge. (This is easy
to verify by induction, hint, hint.) Moreover, since each edge has unit capacity,
the computed flow either saturates (/(e) = 1 ) or avoids (/(e) = 0 ) every edge

^his problem is very different from the stable matching problem we saw in Chapter 4,
because we aren’t trying to make each doctor and hospital as happy as possible.

353

ii. Applications of Flows and Cuts

in G'. Finally, since at most one unit of flow can enter any vertex in U or leave
any vertex in W, the saturated edges from U to W form a matching in G. The
size of this matching is exactly [/ 1 .

Thus, the size of the maximum matching in G is equal to the value of the
maximum flow in G ', and provided we compute the maxflow using augmenting
paths, we can convert the actual maxflow into a maximum matching in 0(E)
time. We can compute the maximum flow in 0(VE) time using either Orlin’s
algorithm or off-the-shelf Ford-Fulkerson.

Figure 11.2. A maximum matching in a bipartite graph G, and the corresponding maximum flow in G 1

A more sophisticated algorithm, proposed by John Hopcroft and Richard
Karp in 1973 , computes maximum matchings in bipartite graphs in O(We)
time.

11.4 Tuple Selection

The bipartite maximum matching problem is the simplest example of a broader
class of problems that I call tuple selection. 2 The input to a tuple selection
problem consists of several finite setsX 1 ,X 2 ,... ,X d , each representing a different
discrete resource. Our task is to select the largest possible set of d-tuples, each
containing exactly one element from each set X u subject to several capacity
constraints of the following form:

• For each index i, Each element x sX t can appear in at most c(x) selected
tuples.

• For each index i, any two elements x e and y e X i+l can appear in at
most c(x,y) selected tuples.

Each of the upper bounds c(x) and c(x,y) is either a (typically small) non¬
negative integer or 00 .

T couldn’t find a standard name for these problems, so I made up my own. These are
sometimes called “assignment problems”, but it’s more common for the phrase “the assignment
problem” to refer to the problem of finding a maximum -weight bipartite matching in an edge-
weighted bipartite graph.

354

ii-4- Tuple Selection

In the maximum-matching problem, we have d = 2 resources, each ele¬
ment x has capacity c(x) = 1, and each pair (x,y) has capacity c(x,y) = 1
or c(x,y) = 0, depending on whether or not xy is an edge in the underlying
bipartite graph.

Because the resources are linearly ordered, and only pairs of objects in
adjacent subsets and X l+] are constrained, 3 the tuple selection problem can
be reduced to a maximum-flow problem in a directed graph G defined as follows:

• G contains a vertex for each element of each set X h as well as a source
vertex s and a target vertex t. Each vertex x (except s and t) has capacity
c(x).

• G contains an edge s->w for each element w e X±, an edge z^t for each
element z e X d , and an edge x^y with capacity c(x,y) for each pair of
elements x e X t and y € X i+1 , for all i. (Optionally, we can omit edges x->y
with c(x,y) = 0.)

Every path from s to t in G corresponds to (or “is”) a d -tuple that we could
select; conversely, every selectable d -tuple that satisfies the stated constraints
corresponds to (or “is”) a path from s to t in G.

Figure 11.3. The flow network for a tuple selection problem.

More generally, let / be an arbitrary feasible integer (s, t)-flow in G. Because
all capacities are integers or 00 , the Flow Decomposition Theorem implies
that / is the sum of \f | paths from s to t, each carrying exactly one unit of
flow. Straightforward definition-chasing implies that the resulting set of tuples
satisfies all the capacity constraints. Conversely, for any set of k tuples that
satisfies the capacity constraints, the sum of the k corresponding paths is a
feasible integer (s, t)-flow with value k.

Thus, we can select the maximum number of tuples that satisfy the given
capacity constraints by computing a maximum (s, t)-flow /* in G. Because all
finite capacities in G are integers, we can assume without loss of generality that
/* is an integer flow, and therefore (by the previous paragraph) corresponds to
a valid set of |/*| tuples.

3 If pairs of objects from even one non-adjacent pair of subsets (X t and Xj where j > i + 1)
are also constrained, the problem becomes NP-hard, by a straightforward reduction from Exact-
3DimensionalMatching. We’ll discuss NP-hardness in the next chapter.

355

ii. Applications of Flows and Cuts

Exam Scheduling

The following “real world” scheduling problem might help clarify our general
reduction.

Sham-Poobanana University has hired you to write an algorithm to schedule
their final exams. There are n different classes, each of which needs to schedule
a final exam in one of r rooms during one of t different time slots. At most one
class’s final exam can be scheduled in each room during each time slot; conversely,
classes cannot be split into multiple rooms or multiple times. Moreover, each
exam must be overseen by one of p proctors. 4 Each proctor can oversee at most
one exam at a time; each proctor is available for only certain time slots; and no
proctor is allowed oversee more than 5 exams total. The input to the scheduling
problem consists of three arrays:

• An integer array E[1.. n] where E[i] is the number of students enrolled in
the ith class.

• An integer array S[l..r], where S[j] is the number of seats in the j th
room. The ith class’s final exam can be held in the j th room if and only if
E[i]<S[j].

• A boolean array A[1.. t, 1.. p] where A[k,l] = True if and only if the (th
proctor is available during the kth time slot. 5

let N = n + r + tp denote the total size of the input. Your job is to design an
algorithm that either schedules a room, a time slot, and a proctor for every
class’s final exam, or correctly reports that no such schedule is possible.

This is a standard tuple-selection problem with four resources: classes,
rooms, time slots, and proctors. To solve this problem, we construct a flow
network G with six types of vertices—a source vertex s', a vertex c ; for each class,
a vertex r } for each room, a vertex t k for each time slot, a vertex p t for each
proctor, and a target vertex t' —and five types of edges, as shown in Figure 11 . 4 :

• An edge s'-h:* with capacity 1 for each class i. (“Each class can hold at most
one final exam.”)

• An edge c i ^r ] with capacity 00 for each class i and room j such that
E[i] < S[j]. (“Class i can hold exams in room j if and only if the room has
enough seats.”) This is the only place where the enrollments E[i] and seat
numbers S[j] are used.

• An edge with capacity 1 for each room j and time slot k. (“At most

one exam can be held in room j at time k.”)

4 or as they are better known outside the US, invigilators

5 Arguably, this information is better represented as a graph, but I thought that would make
the reduction more confusing.

356

11.5. Disjoint-Path Covers

Figure 11.4. A flow network for the exam scheduling problem.

• An edge t k ->p( with capacity 1 for time slot k and proctor t such that
A[£, k] = True. (“A proctor can oversee at most one exam at any time, and
only during times that they are available.”)

• An edge p^t' with capacity 5 for each proctor l. (“Each proctor can oversee
at most 5 exams.”)

(I’m calling the source and target vertices s' and t' instead of s and t only
because the problem statement already uses the variable t to denote the
number of time slots.) Altogether, G has n + r + t+ p + 2 = 0(!V) vertices and
0(nr + rt + tp ) = 0(1V 2 ) edges.

Each path from s' to t' in G represents a unique valid choice of class, room,
time, and proctor for one final exam; specifically, the class fits into the room,
and the proctor is available at that time. Conversely, for each valid choice (class,
room, time, proctor), there is a corresponding path from s' to t' in G. Thus,
we can construct a valid schedule for the maximum possible number of exams
by computing an maximum (s', t'j-flow f* in G, decomposing f* into paths
from s' to t', and then transcribing each path into a class-room-time-proctor
assignment. If |/*| = n, we can return the resulting schedule; otherwise, we
can correctly report that scheduling all n final exams is impossible.

Constructing G from the given input data by brute force takes 0(E) time.
We can compute the maximum flow in 0(VE ) time using either Ford-Fulkerson
(because |/*| < n < V) or Orlin’s algorithm, and we can compute the flow
decomposition in 0(VE ) time. Thus, the overall algorithm runs in 0(VE ) =
0(iV 3 ) time.

11.5 Disjoint-Path Covers

A path cover of a directed graph G is a collection of directed paths in G such that
every vertex of G lies on at least one path. A disjoint- path cover of G is a path
cover such that every vertex of G lies on exactly one path. Every directed graph
has a trivial disjoint-path cover consisting of several paths of length zero, but

357

ii. Applications of Flows and Cuts

that’s boring. Instead, let’s look for disjoint-path covers that contain as few paths
as possible. This problem is NP-hard in general—a graph has a disjoint-path
cover of size 1 if and only if it contains a Hamiltonian path—but there is an
efficient flow-based algorithm for directed acyclic graphs.

To solve this problem for a given directed acyclic graph G = (V,E), we
construct a new bipartite graph G' = ( V', E r ) as follows.

• H contains two vertices v’ and v : for every vertex v of G.

• H contains an undirected edge uM for every directed edge u^v in G.

(If G is represented as an adjacency matrix, then G' is the bipartite graph
represented by the same adjacency matrix!)

Figure 11.5. Reducing minimum disjoint-path cover of a dag to maximum bipartite matching; squares
are flat 1 ' and diamonds are sharp®.

Now I claim that G can be covered by k disjoint paths if and only if the new
graph G' has a matching of size V — k. As usual, we prove the equivalence in
two stages:

<f= Suppose G has a disjoint path cover P with k paths; think of P as a subgraph
of G. Every vertex in P has in-degree either 0 or 1; moreover, there is exactly
one vertex with in-degree 0 in each path in P. It follows that P has exactly
V — k edges. Now define a subset M of the edges of G' as follows:

M := {uM e | iz^veP}.

By definition of disjoint-path cover, every vertex of G has at most one
incoming edge in P and at most one outgoing edge in P. We conclude that
every vertex of G' is incident to at most one edge in M; that is, M is a
matching of size V — k.

=> Suppose G' has a matching M of size V — k. We project M' back to G by
defining a subgraph P = (V, M'), where

M' := {u^v e£ | uMeM}.

By definition of matching, every vertex of G has at most one incoming edge
in P and at most one outgoing edge in P. It follows that P is a collection

358

11.5. Disjoint-Path Covers

of disjoint directed paths in G; since P includes every vertex, P defines an
disjoint path cover with V — k edges. The number of paths in P is equal to
the number of vertices in G that have no incoming edge in M'. We conclude
that P contains exactly k paths.

It follows immediately that we can find a minimum disjoint-path cover in G by
computing a maximum matching in G', using Ford-Fulkerson’s maximum-flow
algorithm, in 0(V'E') = O(VE) time.

Despite its formulation in terms of dags and paths, this is really a maximum
matching problem: We want to match as many vertices as possible to distinct
successors in the graph. The number of paths required to cover the dag is equal
to the number of vertices with no successor. (And of course, every bipartite
maximum matching problem is really a flow problem.)

Minimal Teaching Assignment

Let’s go back to Sham-Poobanana University for another “real-world” scheduling
problem. 6 SPU offers several thousand courses every day. Due to extreme budget
cuts, the university needs to significantly reduce the size of its faculty. However,
because students pay tuition (and the university cannot afford lawyers), the
university must retain enough professors to guarantee that every class advertised
in the course catalog is actually taught. How few professors can SPU get away
with? Each remaining faculty member will be assigned a sequence of classes to
teach on any given day. The classes assigned to each professor must not overlap;
moreover, there must be enough slack in each professor’s schedule for them to
walk from one class to the next. For purposes of this problem, let’s assume that
every professor is capable of teaching every class, and that professors will not
have office hours, lunches, or bathroom breaks. 7

Concretely, suppose there are n classes offered in m different locations. The
input to our problem consists of the following data:

• An array C[l..n] of classes, where each class C[i] has three fields: the
starting time C[i].start, the ending time C[i].end, and the location C[i].loc.

• A two-dimensional array T[1.. m, 1.. m], where T[u, v] is the time required
to walk from location u to location v.

We want to find the minimum number of professors that can collectively teach
every class, such that whenever a professor is assigned to teach two classes i
and j where C[j].start > C[i], start, we actually have

C[j].start > C[i].end +T\_C[i].loc,C[j].loc\.

6 For a somewhat more realistic (and less depressing) formulation of this problem, consider
airplanes and flights, or buses and bus routes, instead of professors and classes.

7 They will, however, be expected to answer student emails as they walk between classes.

359

ii. Applications of Flows and Cuts

We can solve this problem by reducing it to a disjoint-path cover problem as
follows. We construct a dag G = (V,E) whose vertices are classes and whose
edges represent pairs of classes that are scheduled far enough apart to be taught
by the same professor. Specifically, a directed edge i-*j indicates that the same
professor can teach class i and then class j. It is easy to construct this dag in
0 (n 2 ) time by brute force. Then we can find a disjoint-path cover of G using the
matching algorithm described above; each directed path in G represents a legal
class schedule for one professor. The entire algorithm runs in 0 (n 2 + VE) =
0(n 3 ) time. 8

Despite its initial description in terms of intervals and distances, this is
really a maximum matching problem. Specifically, we want to match as many
classes as possible to the next class taught by the same professor. The number of
professors we need is equal to the number of classes with no assigned successor;
each class without an assigned successor is the last class that some professor
teaches.

11.6 Baseball Elimination

Every year millions of American baseball fans eagerly watch their favorite team,
hoping they will win a spot in the playoffs, and ultimately the World Series.
Sadly, most teams are “mathematically eliminated” days or even weeks before the
regular season ends. Often, it is easy to spot when a team is eliminated—they
can’t win enough games to catch up to the current leader in their division.
But sometimes the situation is more subtle. For example, here are the actual
standings from the American League East on August 30, 1996.

Team

Won

Lost

Left

NYY

BAL

BOS

TOR

DET

New York Yankees

75-

59

28

3

8

7

3

Baltimore Orioles

71-

63

28

3

2

7

4

Boston Red Sox

69-

66

27

8

2

0

0

Toronto Blue Jays

63-

72

27

7

7

0

0

Detroit Tigers

49-

86

27

3

4

0

0

Detroit is clearly behind, but some die-hard Tigers fans may hold out hope
that their team can still win. After all, if Detroit wins all 27 of their remaining
games, they will end the season with 76 wins, more than any other team has
now. So as long as every other team loses every game... but that’s not possible,

8 Most American universities schedule ten-minute breaks between classes, under the remark¬
able belief that a normal human being can get from any two classroom on campus to any other
classroom in ten minutes. If we assume that the time interval T[u,v] is identical for all u and v,
this scheduling problem can actually be solved in 0(n log n) time using a simple greedy algorithm.

360

11 . 6 . Baseball Elimination

because some of those other teams still have to play each other. Here is one
complete argument: 9

By winning all of their remaining games, Detroit can finish the season with a record of
76 and 86. If the Yankees win just 2 more games, then they will finish the season with
a 77 and 85 record which would put them ahead of Detroit. So, let's suppose the Tigers
go undefeated for the rest of the season and the Yankees fail to win another game.

The problem with this scenario is that New York still has 8 games left with Boston.

If the Red Sox win all of these games, they will end the season with at least 77 wins
putting them ahead of the Tigers. Thus, the only way for Detroit to even have a chance
of finishing in first place, is for New York to win exactly one of the 8 games with Boston
and lose all their other games. Meanwhile, the Sox must lose all the games they play
against teams other than New York. This puts them in a 3-way tie for first place....

Now let's look at what happens to the Orioles and Blue Jays in our scenario. Balti¬
more has 2 games left with with Boston and 3 with New York. So, if everything happens
as described above, the Orioles will finish with at least 76 wins. So, Detroit can catch
Baltimore only if the Orioles lose all their games to teams other than New York and
Boston. In particular, this means that Baltimore must lose all 7 of its remaining games
with Toronto. The Blue Jays also have 7 games left with the Yankees and we have al¬
ready seen that for Detroit to finish in first place, Toronto must will all of these games.

But if that happens, the Blue Jays will win at least 14 more games giving them at final
record of 77 and 85 or better which means they will finish ahead of the Tigers. So, no
matter what happens from this point in the season on, Detroit can not finish in first
place in the American League East.

There has to be a better way to figure this out!

Here is a more abstract formulation of the problem. Our input consists of
two arrays W[ 1 .. n ] and G[ 1 .. n, 1 .. n], where W[t] is the number of games
team i has already won, and G[i,j ] is the number of upcoming games between
teams i and j. We want to determine whether team n can end the season with
the most wins (possibly tied with other teams). 10

In the mid-1960s, Benjamin Schwartz observed that this question can be
modeled as a maximum flow problem; about 20 years later, Dan Gusfield, Charles
Martel, and David Fernandez-Baca simplified Schwartz’s flow formulation to
a pair selection problem. Specifically, we want to know whether it is possible
to select a winner for each game, so that team n comes in first place. Let
R[i ] = X; G[i, j] denote the number of remaining games for team i. We will
assume that team n wins all R[n] of its remaining games. Then team n can come
in first place if and only if every other team i wins at most W[n]+R[n] — W[t]
of its R[i] remaining games.

Since we want to select a winning team for each game, we start by building
a bipartite graph, whose nodes represent the games and the teams. We have

9 Both the example and this argument are taken from Eli V. Olinick’s web site https:
//S2. smu.edu/~olinick/riot/detroit.html, which is based on Olinick’s joint research with Ilan
Adler, Alan Erera, and Dorit Hochbaum.

1D We assume here that no games end in a tie (true for Major League Baseball games that
affect postseason standing, and mostly true otherwise), and that every game is actually played
(not true for Major League Baseball games during the regular season).

361

ii. Applications of Flows and Cuts

(2) game nodes gy, one for each pair 1 < i < j < n, and n — 1 team nodes t ; ,
one for each 1 < i < n. For each pair i,j, we add edges gy-►£■* and gy->tj
with infinite capacity. We add a source vertex s and edges s->g ; j with capacity
G[i,j ] for each pair i,j. Finally, we add a target node t and edges t;->t with
capacity W[n] — W[i] + R[n] for each team i.

Here is the graph derived from the 1996 American League East standings,
where “team n” is the Detroit Tigers. All unlabeled edges have infinite capacity.

Theorem. Team n can end the season in first place if and only if there is a
feasible flow in this graph that saturates every edge leavings.

Proof: Suppose it is possible for team n to end the season in first place. Then
every team i < n wins at most W[n] + R[n] — W[i] of the remaining games. For
each game between team i and team j that team i wins, add one unit of flow
along the path s-ygij-^t^t. Because there are exactly G[i,j ] games between
teams i and j, every edge leaving s is saturated. Because each team i wins at
most W[n ] +H[n] — W[i] games, the resulting flow is feasible.

Conversely, let / be a feasible flow that saturates every edge out of s.
Suppose team i wins exactly/(gy->t;) games against team j, for all i and j.
Then teams i and j play/(gy^t;)+ /(gy^tj) =/(s^gy) = G[i,j] games,
so every upcoming game is played. Moreover, each team i wins a total of
^■/(gy->tj) = < W[n ] +P[n] — W[i] upcoming games, and therefore

at most W[n] + F[n] games overall. Thus, if team n win all their upcoming
games, they end the season in first place. □

In summary, to decide whether our favorite team can win, we construct the
flow network, compute a maximum flow, and report whether than maximum
flow saturates every edge out of s. For example, in the graph shown above,
the total capacity of the edges leaving s is 27 (because there are 27 remaining
games). On the other hand, the total capacity of the edges entering t is only 26,

362

11 7 - Project Selection

which implies that the maximum flow value is at most 26. We conclude that
Detroit is mathematically eliminated. 11

The flow network has 0 (n 2 ) vertices and 0 (n 2 ) edges, and it can be
constructed in 0 (n 2 ) time. Using Orlin’s algorithm, we can compute the
maximum flow in 0(VE ) = 0(n 4 ) time.

This is not the fastest algorithm for the baseball elimination problem. In
200t, Kevin Wayne proved that one can determine all teams that are mathemat¬
ically eliminated in only 0(n 3 ) time, essentially using a single maximum-flow
computation.

11.7 Project Selection

In our final example, suppose we are given a set of n projects that we could
possibly perform. Some projects cannot be started until certain other projects
are completed. The projects and their dependencies are described by a directed
acyclic graph G whose vertices are the projects, where each edge u->v indicates
that project u cannot be performed before project v. (This is exactly the form of
dependency graphs we considered in Chapter 6.4.) Finally, each project v has
an associated profit $(v) which will be given to us if the project is completed;
some projects have negative profits, which we interpret as positive costs. We can
choose to finish any subset X of the projects that includes all its dependents;
that is, for every project x e X, every project that x depends on is also in X.
Our goal is to find a valid subset of the projects whose total profit is as large as
possible. In particular, if all of the jobs have negative profit, the correct answer
is to do nothing.

Figure 11.6. A dependency graph for a set of eight projects. Diamonds indicate profitable projects;
sguares indicate costly projects. Each edge u^v means n depends on v.

At a high level, our task to partition the projects into two subsets S and T ,
the jobs we Select and the jobs we Turn down. So intuitively, we’d like to model
our problem as a minimum cut problem in a certain graph. But in which graph?
How do we enforce prerequisites? We want to maximize profit, but we only
know how to find minimum cuts. And how do we convert negative profits into
positive capacities?

u We got lucky here; it is possible for a team to be eliminated even if the total capacity of all
edges into t is no smaller than the total capacity of edges out of s.

363

ii. Applications of Flows and Cuts

To transform our given constraint graph G into a flow network G', we add a
source vertex s and a target vertex t to the dependency graph, with an edge
s—>v for every profitable job v (with $(v) > 0), and an edge u^t for every costly
job u (with $(tz) < 0 ). Intuitively, we can think of s as a new job (“Sleep!”) with
profit/cost 0 that we must perform last. We assign capacities to the edges of G'
as follows:

• c(s^v) = $(v) for every profitable job v;

• c(u->t) = — $(iz) for every costly job u ;

• c(u->v ) = oo for every dependency edge u->v.

All edge-capacities are positive, so this is a valid input to the maximum cut
problem.

Now consider an arbitrary (s, t)-cut (S, T) in G'. For any edge u->v in the
original dependency graph, if u e S and veT, then ||S, T[| = oo. Thus, we can
legally select the jobs in S if and only if the capacity of the cut ( S , T) is finite.

Figure 11.7. The flow network for the example dependency graph, along with its minimum cut. The cut
has capacity 13 and P = 15 , so the total profit for the selected jobs is 2 .

In fact, it turns out that cuts with smaller capacity correspond to job
selections with higher profit. Specifically, I claim that selecting the jobs in S
earns a total profit of P — ||S, T ||, where P is the sum of all the positive profits:

P = ^max{0,$(v)} = ^ $(v).

v $(v)>0

We can prove this claim by straightforward definition-chasing, as follows. For
any subset X of projects, we define three values. (Here, as usual, we define
c(u->v ) = 0 when iz->v is not an edge.)

cost(X ) := ^ -$(u) = ^c(iz-At)

ueX ueX

$(u)< 0

yield(X) := ^ $(v) = ^c(s^v)

vex vex

$(v)>0

364

Exercises

profit{X ) := ^ $(v) = yield(X ) — cost(X ).

By definition, P = yield(V ) = yield(S)+yield(T). Because the cut (S, T) has
finite capacity, only edges of the form s->v and u-+t can cross the cut. By
construction, every edge s->v points to a profitable job and each edge u->t points
from a costly job. Thus, ||S, T\\ = cost(S) +yield(T ). We immediately conclude
that P — ||S, T || = yield(S) — cost(S) = profit(S), as claimed.

It follows immediately that we can maximize our total profit by computing a
minimum cut in G'. We can easily construct G' from G in 0 (V +E ) time, and we
can compute the minimum (s, t)-cut in G' in 0 {VE) time using Orlin’s algorithm.
We conclude that the entire project-selection algorithm runs in O(VE) time.

Exercises

1. Let G = (V,E) be a directed graph where for each vertex v, the in-degree
and out-degree of v are equal. Suppose G contains k edge-disjoint paths
from some vertex u to another vertex v. Under these conditions, must G also
contain k edge-disjoint paths from v to u? Give a proof or a counterexample
with explanation.

2. Given an undirected graph G = (V, £), with three vertices u, v, and w,
describe and analyze an algorithm to determine whether there is a path
from u to w that passes through v. [Hint: If G were a directed graph, this
problem would be NP-hard!]

3. Consider a directed graph G = (V, E) with several source vertices s 1 ,s 2 ,... ,s a
and target vertices t la 1 1,..., t T , where no vertex is both a source and a
target. A multi-terminal flow is a function f : E —* R> 0 that satisfies the flow
conservation constraint at every vertex that is neither a source nor a target.
The value \f\ of a multi-terminal flow is the total excess flow out of all the
source vertices:

As usual, we are interested in finding flows with maximum value, subject to
capacity constraints on the edges. (In particular, we don’t care how much
flow moves from any particular source to any particular target.)

(a) Consider the following algorithm for computing multi-terminal flows.
The variables / and f represent flow functions. The subroutine
MaxFlow(G,s, t) solves the standard maximum flow problem with
source s and target t.

365

ii. Applications of Flows and Cuts

MaxMultiFlow(G,s[1

. cr], t[l.. tt]):

/«- 0

for i ^ 1 to cr

((Initialize the flow))

for j <— 1 to T

f <— MaxFlow(G^, s[i], t[j])

/ <—/+/' ((Update the flow}}

return /

Prove that this algorithm correctly computes a maximum multi-terminal
flow in G.

(b) Describe a more efficient algorithm to compute a maximum multi¬
terminal flow in G.

4. The Island of Sodor is home to a large number of towns and villages,
connected by an extensive rail network. Recently, several cases of a deadly
contagious disease (either swine flu or zombies; reports are unclear) have
been reported in the village of Ffarquhar. The controller of the Sodor railway
plans to close down certain railway stations to prevent the disease from
spreading to Tidmouth, his home town. No trains can pass through a closed
station. To minimize expense (and public notice), he wants to close down
as few stations as possible. However, he cannot close the Ffarquhar station,
because that would expose him to the disease, and he cannot close the
Tidmouth station, because then he couldn’t visit his favorite pub.

Describe and analyze an algorithm to find the minimum number of
stations that must be closed to block all rail travel from Ffarquhar to
Tidmouth. The Sodor rail network is represented by an undirected graph,
with a vertex for each station and an edge for each rail connection between
two stations. Two special vertices / and t represent the stations in Ffarquhar
and Tidmouth.

For example, given the following input graph, your algorithm should
return the integer 2.

5. Ann x n grid is an undirected graph with n 2 vertices organized into n rows
and n columns. We denote the vertex in the ith row and the jth column
by (i,j). Every vertex (i, j) has exactly four neighbors (i — 1 , j), (i + 1 , j),
(i, j — 1), and (i, j + 1), except the boundary vertices, for which i = 1, i = n,
j = 1, or j = n.

366

Exercises

Let (x 1; y 1 ), (x 2 ,y 2 ), •••> ( x m> Ym) be distinct vertices, called terminals,
in the n x n grid. The escape problem is to determine whether there are m
vertex-disjoint paths in the grid that connect the terminals to any m distinct
boundary vertices.

Figure 11.8. A positive instance of the escape problem, and its solution.

(a) Describe and analyze an efficient algorithm to solve the escape problem.
The running time of your algorithm should be a small polynomial function
of n.

(b) Now suppose the input to the escape problem consists of a single integer
n and the list of m terminal vertices. If m is very small, the previous
running time is actually exponential in the input size! Describe and
analyze an algorithm to solve the escape problem in time polynomial in
m.

r (c) Modify the previous algorithm to output an explicit description of the
escape paths (if they exist), still in time polynomial in m.

6 . The UIUC Computer Science Department is installing a mini-golf course
in the basement of Siebel Center! The playing field is a closed polygon
bounded by m horizontal and vertical line segments, meeting at right angles.
The course has n starting points and n holes, in one-to-one correspondence.
It is always possible hit the ball along a straight line directly from each
starting point to the corresponding hole, without touching the boundary
of the playing field. (Players are not allowed to bounce golf balls off the
walls; too much glass.) The n starting points and n holes are all at distinct
locations.

Sadly, the architect’s computer crashed just as construction was about to
begin. Thanks to the herculean efforts of their sysadmins, they were able to
recover the locations of the starting points and the holes, but all information
about which starting points correspond to which holes was lost!

Describe and analyze an algorithm to compute a one-to-one correspon¬
dence between the starting points and the holes that meets the straight-line
requirement, or to report that no such correspondence exists. The input

367

ii. Applications of Flows and Cuts

Figure n.g. A mini-golf course with five starting points (*) and holes (o), and a legal correspondence
between them.

consists of the x- and y-coordinates of the m corners of the playing field, the
n starting points, and the n holes. Assume you can determine in constant
time whether two line segments intersect, given the x- and y-coordinates
of their endpoints.

7. A cycle cover of a given directed graph G = (V,£) is a set of vertex-disjoint
cycles that cover every vertex in G. Describe and analyze an efficient
algorithm to find a cycle cover for a given graph, or correctly report that no
cycle cover exists. [Hint: Use bipartite matching!]

8. Suppose you are given antxn checkerboard with some of the squares
deleted. You have a large set of dominos, just the right size to cover
two squares of the checkerboard. Describe and analyze an algorithm to
determine whether one tile the board with dominos—each domino must
cover exactly two undeleted squares, and each undeleted square must be
covered by exactly one domino.

rn

- ( 1

0

PI

LI

J U

1 )

- (

□ Q

1 1

PI PI

djL

- 1

1

TQI

- 1

11 1

Figure 11.10. Covering a partial checkerboard with dominos.

Your input is a boolean array De?eted[l.. n, 1 .. n], where Deleted[i, j ] =
True if and only if the square in row i and column j has been deleted. Your
output is a single boolean; you do not have to compute the actual placement
of dominos. For example, for the board shown above, your algorithm should
return True.

368

Exercises

9. Suppose we are given an n x n square grid, some of whose squares are
colored black and the rest white. Describe and analyze an algorithm to
determine whether tokens can be placed on the grid so that

• every token is on a white square;

• every row of the grid contains exactly one token; and

• every column of the grid contains exactly one token.

Your input is a two dimensional array IsWhite[l .. n, 1 .. n] of booleans,
indicating which squares are white. Your output is a single boolean. For
example, given the grid above as input, your algorithm should return True.

Figure 11.11. Marking every row and column in a grid.

10. Suppose we are given a set of boxes, each specified by their height, width,
and depth in centimeters. All three side lengths of every box lie strictly
between 10cm and 20cm. As you should expect, one box can be placed
inside another if the smaller box can be rotated so that its height, width,
and depth are respectively smaller than the height, width, and depth of the
larger box. Boxes can be nested recursively. Call a box is visible if it is not
inside another box.

Describe and analyze an algorithm to nest the boxes so that the number
of visible boxes is as small as possible.

11. Suppose we are given an nxn grid, some of whose cells are marked; the grid
is represented by an array M[ 1 .. n, 1 .. n] of booleans, where M[i,j ] = True
if and only if cell (i, j) is marked. A monotone path through the grid starts
at the top-left cell, moves only right or down at each step, and ends at
the bottom-right cell. Our goal is to cover the marked cells with as few
monotone paths as possible.

(a) Describe an algorithm to find a monotone path that covers the largest
number of marked cells.

(b) There is a natural greedy heuristic to find a small cover by monotone
paths: If there are any marked cells, find a monotone path n that covers

369

ii. Applications of Flows and Cuts

the largest number of marked cells, unmark any cells covered by n those
marked cells, and recurse. Show that this algorithm does not always
compute an optimal solution.

(c) Describe and analyze an efficient algorithm to compute the smallest set
of monotone paths that covers every marked cell.

12. The Faculty Senate at Sham-Poobanana University has decided to convene a
committee to determine whether Uncle Gabby, Professor Bobo Cornelius,
or Mofo the Psychic Gorilla should replace the recently disgraced Baron
Factotum as the new official mascot symbol of SPU’s athletic teams (The
Fighting Pooh-bahs). Exactly one faculty member must be chosen from each
academic department to serve on this committee. Some faculty members
have appointments in multiple departments, but each committee member
can represent only one department. For example, if Prof. Blagojevich is
affiliated with both the Department of Corruption and the Department of
Stupidity, and he is chosen as the Stupidity representative, then someone
else must represent Corruption. Finally, University policy requires that
every faculty committee must contain exactly the same number of assistant
professors, associate professors, and full professors. Fortunately, the number
of departments is a multiple of 3.

Describe and analyze an algorithm to choose a subset of the SPU faculty
to staff The Post-Factotum Simian Mascot Symbol Committee, or correctly
report that no valid committee is possible. Your input is a bipartite graph
indicating which professors belong to which departments; each professor
vertex is labeled with that professor’s rank (assistant, associate, or full).

13. The Department of Commuter Silence at Sham-Poobanana University has
a flexible curriculum with a complex set of graduation requirements. The
department offers n different courses, and there are m different requirements.
Each requirement specifies a subset of the n courses and the number of
courses that must be taken from that subset. The subsets for different
requirements may overlap, but each course can only be used to satisfy at
most one requirement.

370

Exercises

For example, suppose there are n = 5 courses A,B,C,D,E and m = 2
graduation requirements:

• You must take at least 2 courses from the subset {A, B, C }.

• You must take at least 2 courses from the subset {C, D,E}.

Then a student who has only taken courses B, C, D cannot graduate, but a
student who has taken either A,B,C,D or B, C, D, E can graduate.

Describe and analyze an algorithm to determine whether a given student
can graduate. The input to your algorithm is the list of m requirements
(each specifying a subset of the n courses and the number of courses that
must be taken from that subset) and the list of courses the student has taken.

14. You’re organizing the First Annual SPU Commuter Silence 72-Hour Dance
Exchange, to be held all day Friday, Saturday, and Sunday. Several 30-minute
sets of music will be played during the event, and a large number of DJs
have applied to perform. You need to hire DJs according to the following
constraints.

• Exactly k sets of music must be played each day, and thus 3 k sets
altogether.

• Each set must be played by a single DJ in a consistent music genre
(ambient, bubblegum, dubstep, horrorcore, K-pop, Kwaito, mariachi,
straight-ahead jazz, trip-hop, Nashville country, parapara, ska, ... ).

• Each genre must be played at most once per day.

• Each candidate DJ has given you a list of genres they are willing to play.

• Each DJ can play at most three sets during the entire event.

Suppose there are n candidate DJs and g different musical genres available.
Describe and analyze an efficient algorithm that either assigns a DJ and a
genre to each of the 3 k sets, or correctly reports that no such assignment is
possible.

15. Suppose you are running a web site that is visited by the same set of people
every day. Each visitor claims membership in one or more demographic
groups; for example, a visitor might describe himself as male, 40-50 years
old, a father, a resident of Illinois, an academic, a blogger, and a fan of Joss
Whedon. 12 Your site is supported by advertisers. Each advertiser has told
you which demographic groups should see its ads and how many of its ads
you must show each day. Altogether, there are n visitors, k demographic
groups, and m advertisers.

12 Har har har! Mine is an evil laugh! Now die!

371

ii. Applications of Flows and Cuts

Describe an efficient algorithm to determine, given all the data described
in the previous paragraph, whether you can show each visitor exactly one
ad per day, so that every advertiser has its desired number of ads displayed,
and every ad is seen by someone in an appropriate demographic group.

16. Suppose we are given an array A[ 1 .. m][l.. n] of non-negative real numbers.
We want to round A to an integer matrix, by replacing each entry x in A
with either [xj or [x], without changing the sum of entries in any row or
column of A. For example:

1.2

3.4

2.4

1

4

2

3.9

4.0

2.1

—»

4

4

2

7.9

1.6

0.5

8

1

1

(a) Describe and analyze an efficient algorithm that either rounds A in this
fashion, or reports correctly that no such rounding is possible.

(b) Prove that a legal rounding is possible if and only if the sum of entries
in each row is an integer, and the sum of entries in each column is an
integer. In other words, prove that either your algorithm from part (a)
returns a legal rounding, or a legal rounding is obviously impossible.

*(c) Suppose we are guaranteed that none of the entries in the input matrix
A is an integer. Describe and analyze an even faster algorithm that either
rounds A or reports correctly that no such rounding is possible. For full
credit, your algorithm must run in O(mn') time. [Hint: Don’t use flows.]

17. Ad-hoc networks are made up of low-powered wireless devices. In prin¬
ciple 13 , these networks can be used on battlefields, in regions that have
recently suffered from natural disasters, and in other hard-to-reach areas.
The idea is that a large collection of cheap, simple devices could be distrib¬
uted through the area of interest (for example, by dropping them from an
airplane); the devices would then automatically configure themselves into a
functioning wireless network.

These devices can communicate only within a limited range. We assume
all the devices are identical; there is a distance D such that two devices can
communicate if and only if the distance between them is at most D.

We would like our ad-hoc network to be reliable, but because the devices
are cheap and low-powered, they frequently fail. If a device detects that
it is likely to fail, it should transmit its information to some other backup
device within its communication range. We require each device x to have k

13 but not so much in practice

372

Exercises

potential backup devices, all within distance D of x; we call these k devices
the backup set of x. Also, we do not want any device to be in the backup
set of too many other devices; otherwise, a single failure might affect a large
fraction of the network.

So suppose we are given the communication radius D, parameters b
and k, and an array d[l ..n, 1 ..n] of distances, where d[i,j ] is the distance
between device i and device j. Describe an algorithm that either computes
a backup set of size k for each of the n devices, such that no device appears
in more than b backup sets, or reports (correctly) that no good collection of
backup sets exists.

18. Faced with the threat of brutally severe budget cuts, Potemkin University
has decided to hire actors to sit in classes as “students”, to ensure that
every class they offer is completely full. Because actors are expensive, the
university wants to hire as few of them as possible.

Building on their previous leadership experience at the now-defunct
Sham-Poobanana University, the administrators at Potemkin have given
you a directed acyclic graph G = (V,E), whose vertices represent classes,
and where each edge i-*j indicates that the same “student” can attend
class i and then later attend class j. In addition, you are also given an array
cap[l.. V ] listing the maximum number of “students” who can take each
class. Describe an analyze an algorithm to compute the minimum number
of “students” that would allow every class to be filled to capacity.

19. Quentin, Alice, and the other Brakebills Physical Kids are planning an
excursion through the Neitherlands to Fillory The Neitherlands is a vast,
deserted city composed of several plazas, each containing a single fountain
that can magically transport people to a different world. Adjacent plazas are
connected by gates, which have been cursed by the Beast. The gates between
plazas are open only for five minutes every hour, all simultaneously—from
12:00 to 12:05, then from 1:00 to 1:05, and so on—and are otherwise locked.
During those five minutes, if more than one person passes through any single
gate, the Beast will detect their presence. 14 Moreover, anyone attempting
to open a locked gate, or attempting to pass through more than one gate
within the same five-minute period will turn into a niffin. 15 However, any
number of people can safely pass through different gates at the same time
and/or pass through the same gate at different times.

14 This is very bad.
15 This is very very bad.

373

ii. Applications of Flows and Cuts

You are given a map of the Neitherlands, which is a graph G with a
vertex for each fountain and an edge for each gate, with the fountains to
Earth and Fillory clearly marked.

(a) Suppose you are also given a positive integer h. Describe and analyze
an algorithm to compute the maximum number of people that can walk
from the Earth fountain to the Fillory fountain in at most h hours—that
is, after the gates have opened at most h times—without anyone alerting
the Beast or turning into a niffin. The running time of your algorithm
should depend on h. [Hint: Build a different graph.]

* v (b) Describe an analyze an algorithm for part (a) whose running time is
polynomial in V and E, with no dependence on h.

(c) On the other hand, suppose you are also given an integer k. Describe
and analyze an algorithm to compute the minimum number of hours that
allow k people to walk from the Earth fountain to the Fillory fountain,
without anyone alerting the Beast or turning into a niffin. [Hint: Use
part (a).]

*20. Let G = (LUk,E) bea bipartite graph, whose left vertices L are indexed
l\,l 2 , and whose right vertices are indexed r 1 ,r 2 ,... ,r n . A matching

M in G is non-crossing if, for every pair of edges i r ; and t^ry in M, we
have i < i' if and only if j < j'.

(a) Describe and analyze an algorithm to find the largest non-crossing
matching in G. [Hint: This is not really a flow problem.]

(b) Describe and analyze an algorithm to find the smallest number of non¬
crossing matchings M 1 ,M 2 , ■ ■ ■, M k such that each edge in G is in exactly
one matching M ; . [Hint: This is really a flow problem.]

*21. Let G = (LUk,£) bea bipartite graph, whose left vertices L are indexed
ti,l 2 ,...,l n in some arbitrary order.

(a) A matching M in G is dense if there are no consecutive unmatched
vertices in L; that is, for each index i, either or l i+1 is incident to an
edge in M. Describe an algorithm to determine whether G has a dense
matching.

(b) A matching M in G is sparse if there are no consecutive matched vertices
in L; that is, for each index i, either l t or £ i+1 is not incident to an
edge in M. (In particular, the empty matching is sparse.) Describe an
algorithm to find the largest sparse matching in G.

(c) A matching M in G is palindromic if, for every index i, either (, and
i n - i+ 1 are both incident to edges in M, or neither nor f n _ i+1 is incident

374

Exercises

to an edge in M. (In particular, the empty matching is palindromic.)

Describe an algorithm to find the largest palindromic matching in G.

None of these problems restrict which vertices in R are matched or un¬
matched.

*22. A rooted tree is a directed acyclic graph, in which every vertex has exactly
one incoming edge, except for the root, which has no incoming edges.
Equivalently, a rooted tree consists of a root vertex, which has edges pointing
to the roots of zero or more smaller rooted trees. Describe an efficient
algorithm to compute, given two rooted trees A and B, the largest rooted
tree that is isomorphic to both a subgraph of A and a subgraph of B. More
briefly, describe an algorithm to find the largest common subtree of two
rooted trees.

[Hint: This would be a relatively straightforward dynamic programming
problem if either every node had 0(1) children or the children of each node
were ordered from left to right. But for unordered trees with large degree,
you need another technique to combine recursive subproblems efficiently.]

375

[I]n his short and broken treatise he provides an eternal example—not of laws, or
even of method, for there is no method except to be very intelligent, but of
intelligence itself swiftly operating the analysis of sensation to the point of
principle and definition.

- T. S. Eliot on Aristotle, "The Perfect Critic", The Sacred Wood (1921)

The nice thing about standards is that you have so many to choose from ;
furthermore, if you do not like any of them, you can just wait for next year's model.

— Andrew S. Tannenbaum, Computer Networks (1981)

It is a rare mind indeed that can render the hitherto non-existent blindingly obvious.
The cry "I could have thought of that" is a very popular and misleading one, for the
fact is that they didn't, and a very significant and revealing fact it is too.

- Dirk Gently to Richard McDuff
in Douglas Adams' Dirk Gently's Holistic Detective Agency (1987)

If a problem has no solution, it may not be a problem, but a fact -
not to be solved, but to be coped with over time.

— Shimon Peres, as quoted by David Rumsfeld, Rumsfeld's Rules (2001)

NP-Hardness

12.1 A Game You Can't Win

Imagine that a salesman in a red suit, who looks suspiciously like Tom Waits,
presents you with a black steel box with n binary switches on the front and a
light bulb on the top. The salesman tells you that the state of the light bulb is
controlled by a complex boolean circuit — a collection of And, Or, and Not gates
connected by wires, with one input wire for each switch and a single output
wire for the light bulb. He then asks you the following question: Is there a
way to set the switches so that the light bulb turns on? If you can answer this
question correctly, he will give you the box and a million billion trillion dollars;
if you answer incorrectly, or if you die without answering at all, he will take
your soul.

As far as you can tell, the Adversary hasn’t connected the switches to the
light bulb at all, so no matter how you set the switches, the light bulb will stay
off. If you declare that it is possible to turn on the light, the Adversary will open

377

12. NP-Hardness

Figure 12.1 . An And gate, an Or gate, and a Not gate.

Figure 12.2. A boolean circuit. Inputs enter from the left, and the output leaves to the right.

the box and reveal that there is no circuit at all. But if you declare that it is
not possible to turn on the light, before testing all 2 n settings, the Adversary
will magically create a circuit inside the box that turns on the light if and only
if the switches are in one of the settings you haven’t tested, and then flip the
switches to that setting, turning on the light. (You can’t detect the Adversary’s
cheating, because you can’t see inside the box until the end.) The only way
to provably answer the Adversary’s question correctly is to try all 2 n possible
settings. You quickly realize that this will take far longer than you expect to
live, so you gracefully decline the Adversary’s offer.

The Adversary smiles and says, in a growl like Heath Ledger’s Joker after
smoking a carton of Marlboros, “Ah, yes, of course, you have no reason to trust
me. But perhaps I can set your mind at ease.” He hands you a large roll of
parchment—which you hope was made from sheep skin—with a circuit diagram
drawn (or perhaps tattooed) on it. “Here are the complete plans for the circuit
inside the box. Feel free to poke around inside the box to make sure the plans
are correct. Or build your own circuit from these plans. Or write a computer
program to simulate the circuit. Whatever you like. If you discover that the
plans don’t match the actual circuit in the box, you win the trillion bucks.” A few
spot checks convince you that the plans have no obvious flaws; subtle cheating
appears to be impossible.

But you should still decline the Adversary’s “generous” offer. The problem
that the Adversary is posing is called circuit satisfiability or CircuitSat\ Given
a boolean circuit, is there is a set of inputs that makes the circuit output True,
or conversely, whether the circuit always outputs False. For any particular input
setting, we can calculate the output of the circuit in polynomial (actually, linear')
time using depth-first-search. But nobody knows how to solve CircuitSat faster
than just trying all 2 n possible inputs to the circuit by brute force, which requires

378

12 . 2 . P versus NP

exponential time. Admittedly, nobody has actually formally proved that we can’t
beat brute force—maybe, just maybe, there’s a clever algorithm that just hasn’t
been discovered yet—but nobody has actually formally proved that anti-gravity
unicorns don’t exist, either. For all practical purposes, it’s safe to assume that
there is no fast algorithm for CircuitSat.

You tell the salesman no. He smiles and says, “You’re smarter than you look,
kid,” and then flies away on his anti-gravity unicorn.

12.2 P versus NP

A minimal requirement for an algorithm to be considered “efficient” is that its
running time is bounded by a polynomial function of the input size: 0(n c ) for
some constant c, where n is the size of the input. 1 Researchers recognized
early on that not all problems can be solved this quickly, but had a hard time
figuring out exactly which ones could and which ones couldn’t. There are
several so-called NP-hard problems, which most people believe cannot be solved
in polynomial time, even though nobody can prove a super-polynomial lower
bound.

A decision problem is a problem whose output is a single boolean value: Yes
or No. Let me define three classes of decision problems:

• P is the set of decision problems that can be solved in polynomial time.
Intuitively, P is the set of problems that can be solved quickly.

• NP is the set of decision problems with the following property: If the answer
is Yes, then there is a proof of this fact that can be checked in polynomial
time. Intuitively, NP is the set of decision problems where we can verify a
Yes answer quickly if we have the solution in front of us.

• co-NP is essentially the opposite of NP. If the answer to a problem in co-NP
is No, then there is a proof of this fact that can be checked in polynomial
time.

For example, the circuit satisfiability problem is in NP. If a given boolean circuit
is satisfiable, then any set of m input values that produces True output is a
proof that the circuit is satisfiable; we can check the proof by evaluating the
circuit in polynomial time. It is widely believed that circuit satisfiability is not
in P or in co-NP, but nobody actually knows.

Every decision problem in P is also in NP. If a problem is in P, we can verify
Yes answers in polynomial time recomputing the answer from scratch! Similarly,
every problem in P is also in co-NP.

'This notion of efficiency was independently formalized by Alan Cobham in 1965, Jack
Edmonds in 1965, and Michael Rabin in 1966, although similar notions were considered more
than a decade earlier by Kurt Godel, John Nash, and John von Neumann.

379

12. NP-Hardness

Perhaps the single most important unanswered question in theoretical
computer science—if not all of computer science—if not all of science —is
whether the complexity classes P and NP are actually different. Intuitively,
it seems obvious to most people that P 7^ NP; the homeworks and exams in
your algorithms and data structures classes have (I hope) convinced you that
problems can be incredibly hard to solve, even when the solutions are simple
in retrospect. It’s completely obvious; of course solving problems from scratch
is harder than just checking that a solution is correct. We can reasonably
accept—and most algorithm designers do accept—the statement “P f NP” as a
law of nature, similar to other laws of nature like Maxwell’s equations, general
relativity, and the sun rising tomorrow morning that are strongly supported by
evidence, but have no mathematical proof.

But if we’re being mathematically rigorous, we have to admit that nobody
knows how to prove that that P / NP. In fact, there has been little or no real
progress toward a proof for decades. 2 The Clay Mathematics Institute lists
P versus NP as the first of its seven Millennium Prize Problems, offering a
$1,000,000 reward for its solution. And yes, in fact, several people have lost
their souls, or at least their sanity, attempting to solve this problem.

A more subtle but still open question is whether the complexity classes NP
and co-NP are different. Even if we can verify every Yes answer quickly, there’s
no reason to believe we can also verify No answers quickly. For example, as far
as we know, there is no short proof that a boolean circuit is not satisfiable. It is
generally believed that NP / co-NP, but again, nobody knows how to prove it.

Figure 12.3. What we think the world looks like.

12.3 NP-hard, NP-easy, and NP-complete

A problem II is NP-hard if a polynomial-time algorithm for II would imply a
polynomial-time algorithm for every problem in NP. In other words:

n is NP-hard If II can be solved in polynomial time, then P=NP

2 Perhaps the most significant progress has taken the form of barrier results, which imply that
entire avenues of attack are doomed to fail. In a very real sense, not only do we have no idea
how to prove P f NP, but we can actually prove that we have no idea how to prove P f NP!

380

12.3- NP-hard, NP-easy, and NP-complete

Intuitively, if we could solve one particular NP-hard problem quickly, then we
could quickly solve any problem whose solution is easy to understand, using
the solution to that one special problem as a subroutine. NP-hard problems are
at least as hard as every problem in NP.

Finally, a problem is NP-complete if it is both NP-hard and an element of
NP (or “NP-easy”). Informally, NP-complete problems are the hardest problems
in NP. A polynomial-time algorithm for even one NP-complete problem would
immediately imply a polynomial-time algorithm for every NP-complete problem.
Literally thousands of problems have been shown to be NP-complete, so a
polynomial-time algorithm for one (and therefore all) of them seems incredibly
unlikely.

Calling a problem NP-hard is like saying “If I own a dog, then it can speak
fluent English.” You probably don’t know whether or not I own a dog, but I bet
you’re pretty sure that I don’t own a talking dog. Nobody has a mathematical
proof that dogs can’t speak English—the fact that no one has ever heard a dog
speak English is evidence, as are the hundreds of examinations of dogs that
lacked the proper mouth shape and brainpower, but mere evidence is not a
mathematical proof. Nevertheless, no sane person would believe me if I said I
owned a dog that spoke fluent English. So the statement “If I own a dog, then it
can speak fluent English” has a natural corollary: No one in their right mind
should believe that I own a dog! Similarly, if a problem is NP-hard, no one in
their right mind should believe it can be solved in polynomial time.

NP-complete

Figure 12.4. More of what we think the world looks like.

It is not obvious that any problems are NP-hard. The following remarkable
theorem was first published by Steve Cook in 1971 and independently by
Leonid Levin in 1973. 3 I won’t even sketch the proof here, because I’ve been
(deliberately) vague about the definitions; interested readers find a proof in my
lecture notes on nondeterministic Turing machines.

3 Levin first reported his results at seminars in Moscow in 1971, while still a PhD student. News
of Cook’s result did not reach the Soviet Union until at least 1973, after Levin’s announcement of
his results had been published; in accordance with Stigler’s Law, this result is often called “Cook’s
Theorem”. Levin was denied his PhD at Moscow University for political reasons; he emigrated to
the US in 1978 and earned a PhD at MIT a year later. Cook was denied tenure at Berkeley in
1970, just one year before publishing his seminal paper; he (but not Levin) later won the Turing
award for his proof.

381

12. NP-Hardness

The Cook-Levin Theorem. Circuit satisfiability is NP-hard.

*1 2.4 Formal Definitions (HC SVNT DRACONES)

Formally, the complexity classes P, NP, and co-NP are defined in terms of
languages and Turing machines. A language is a set of strings over some finite
alphabet E; without loss of generality, we can assume that E = { 0 , 1 }. A
Turing machine is a very restrictive type of computer whose precise definition
are surprisingly unimportant. P is the set of languages that can be decided
in Polynomial time by a deterministic single-tape Turing machine. Similarly,
NP is the set of all languages that can be decided in polynomial time by a
nondeterministic Turing machine; NP is an abbreviation for IVondcterministic
Polynomial-time.

The requirement of polynomial time is sufficiently crude that we do not
have to specify the precise form of Turing machine (number of tapes, number
of heads, number of tracks, size of the tape alphabet, and so on). In fact, any
algorithm that runs on a random-access machine 4 in T(n) time can be simulated
by a single-tape, single-track, single-head Turing machine that runs in 0 (T(n) 4 * )
time. This simulation result allows us to argue formally about computational
complexity in terms of standard high-level programming constructs like for-
loops and recursion, instead of describing everything directly in terms of Turing
machines.

Formally, a problem n is NP-hard if and only if, for every language Yl' e NP,
there is a polynomial-time Turing reduction from n 7 to II. A Turing reduction
just means a reduction that can be executed on a Turing machine; that is, a
Turing machine M that can solve II 7 using another Turing machine M 7 for Id
as a black-box subroutine. Turing reductions are also called oracle reductions;
polynomial-time Turing reductions are also called Cook reductions.

Researchers in complexity theory prefer to define NP-hardness in terms of
polynomial-time many-one reductions, which are also called Kai~p reductions.
A many-one reduction from one language L' c E* to another language L c E* is
an function / : E* —» E* such that x € L' if and only if f(x) e L. Then we can

4 Random-access machines are a model of computation that more faithfully models physical

computers. A standard random-access machine has unbounded random-access memory, modeled

as an unbounded array M[ 0 .. oo] where each address M[i] holds a single w-bit integer, for

some fixed integer w, and can read to or write from any memory addresses in constant time.
RAM algorithms are formally written in assembly-like language, using instructions like ADD i,j,k
(meaning “M[i] <— M[j] + M[k]”), INDIR i,j (meaning “M[i] <— M[M[j]]”), and IF 0 GOTO i,t
(meaning “if M[i] = 0 , go to line £”). The precise instruction set is surprisingly irrelevant. By
definition, each instruction executes in unit time. In practice, RAM algorithms can be faithfully
described using higher-level pseudocode, as long as we’re careful about arithmetic precision.

382

12.5. Reductions and Sat

define a language L to be NP-hard if and only if, for any language L' € NP, there
is a many-one reduction from L' to L that can be computed in polynomial time.

Every Karp reduction “is” a Cook reduction, but not vice versa. Specifically,
any Karp reduction from one decision problem n to another decision Eh is
equivalent to transforming the input to n into the input for Yl', invoking an
oracle (that is, a subroutine) for Eh, and then returning the answer verbatim.
However, as far as we know, not every Cook reduction can be simulated by a
Karp reduction.

Complexity theorists prefer Karp reductions primarily because NP is closed
under Karp reductions, but is not closed under Cook reductions (unless NP=co-
NP, which is considered unlikely). There are natural problems that are (1)
NP-hard with respect to Cook reductions, but (2) NP-hard with respect to Karp
reductions only if P=NP. One trivial example is of such a problem is UnSat:
Given a boolean formula, is it always false ? On the other hand, many-one
reductions apply only to decision problems (or more formally, to languages);
formally, no optimization or construction problem is Karp-NP-hard.

To make things even more confusing, both Cook and Karp originally defined
NP-hardness in terms of logarithmic-space reductions. Every logarithmic-space
reduction is a polynomial-time reduction, but (as far as we know) not vice
versa. It is an open question whether relaxing the set of allowed (Cook or
Karp) reductions from logarithmic-space to polynomial-time changes the set of
NP-hard problems.

Fortunately, none of these subtleties raise their ugly heads in practice—in
particular, every algorithmic reduction described in these notes can be formalized
as a logarithmic-space many-one reduction—so you can wake up now.

12.5 Reductions and Sat

To prove that any problem other than Circuit satisfiability is NP-hard, we use a
reduction argument. Reducing problem A to another problem B means describing
an algorithm to solve problem A under the assumption that an algorithm for
problem B already exists. You’re already used to doing reductions, only you
probably call it something else, like writing subroutines or utility functions, or
modular programming. To prove something is NP-hard, we describe a similar
transformation between problems, but not in the direction that most people
expect.

You should tattoo the following rule of onto the back of your hand, right
next to your mom’s birthday and the actual rules of Monopoly . 5

5 If a player lands on an available property and declines (or is unable) to buy it, that property
is immediately auctioned off to the highest bidder; the player who originally declined the property
may bid, and bids may be arbitrarily higher or lower than the list price. Players in Jail can still

383

12. NP-Hardness

To prove that problem A is NP-hard,
reduce a known NP-hard problem to A.

In other words, to prove that your problem is hard, you need to describe an
algorithm to solve a different problem, which you already know is hard, using a
magical mystery algorithm for your problem as a subroutine. The essential logic
is a proof by contradiction. The reduction implies that if your problem were
easy, then the other problem would be easy, which it ain’t. Equivalently, since
you know the other problem is hard, the reduction implies that your problem
must also be hard.

As a canonical example, consider the formula satisfiability problem, usually
just called Sat. The input to Sat is a boolean formula like

(a V b V c V d) ((b A c) V (a => d) V (c a A b)),

and the question is whether it is possible to assign boolean values to the variables
a, b, c,... so that the entire formula evaluates to True.

To prove that Sat is NP-hard, we need to give a reduction from a known
NP-hard problem. The only problem we know is NP-hard so far is CircuitSat,
so let’s start there.

Let K be an arbitrary boolean circuit. We can transform K into a boolean
formula $ by creating new output variables for each gate, and then just writing
down the list of gates separated by Ands. For example, our example circuit
would be transformed into a formula as follows:

Oh = AA xfi) A (y 2 = x 4 ) A (y 3 = x 3 A y 2 ) A (y 4 = y 3 V x 2 ) A
Os = *2) A (y 6 = 3qj) A (y 7 = y 3 V y 5 ) A (z = y 4 A y 7 A y 6 ) A z

buy and sell property, buy and sell houses and hotels, and collect rent. The game has 32 houses
and 12 hotels; once they’re gone, they’re gone. In particular, if all houses are already on the
board, you cannot downgrade a hotel to four houses; you must raze all the hotels in the group
to the ground. Players can sell or exchange undeveloped properties with each other, but cannot
sell property back to the bank; on the other hand, players can sell buildings to the bank (at half
price), but cannot sell or exchange buildings with each other. All penalties are paid directly to
the bank. A player landing on Free Parking does not win anything. A player landing on Go gets
exactly $200, no more. Railroads are not magic transporters. Finally, Jeff always gets the boot.
No, not the T-Rex or the penguin—the boot, dammit.

384

12.5. Reductions and Sat

Now we claim that the original circuit K is satisfiable if and only if the
resulting formula 4> is satisfiable. Like every other “if and only if” statement, we
prove this claim in two steps:

=> Given a set of inputs that satisfy the circuit K, we can obtain a satisfying
assignment for the formula <f> by computing the output of every gate in K.

<t= Given a satisfying assignment for the formula <f>, we can obtain a satisfying
input the circuit by simply ignoring the internal gate variables y l and the
output variable z.

The entire transformation from circuit to formula can be carried out in linear
time. Moreover, the size of the resulting formula is at most a constant factor
larger than any reasonable representation of the circuit.

Now suppose, for the sake of argument, there is an algorithm that can
determine in polynomial time whether a given boolean formula is satisfiable.
Then given any boolean circuit K, we can decide whether K is satisfiable by first
transforming K into a boolean formula <f> as described above, and then asking
our magical mystery Sat algorithm whether $ is satisfiable, as suggested by the
following cartoon. Each box represents an algorithm. The red box on the left is
the transformation subroutine. The box on the right is the magic Sat algorithm.
It must be magic, because it has a rainbow on it. 6

If you prefer magic pseudocode to magic boxes:

CircuitSat(JC):

transcribe K into a boolean formula $
return Sat(4>) (( **MAGIC* ))

Transcribing K into <I> requires only polynomial time (in fact, only linear time,
but whatever), so the entire CircuitSat algorithm also runs in polynomial time.

^CircuitSat(^) < O(n) + T SAT (0(n))

We conclude that any polynomial-time algorithm for Sat would give us a
polynomial-time algorithm for CircuitSat, which in turn would imply P=NP.
So Sat is NP-hard!

6 Kate Erickson, personal communication, 2011. For those of you reading black-and-white
printed copies: Yes, that round thing is a rainbow.

385

12. NP-Hardness

i 2.6 3Sat (from Sat)

A special case of Sat that is particularly useful in proving NP-hardness results is
called 3CNF-SAT or more often just 3SAT.

A boolean formula is in conjunctive normal form (CNF) if it is a conjunction
(and) of several clauses, each of which is the disjunction (or) of several literals,
each of which is either a variable or its negation. For example:

clause

,--V

(a V b V c V d) A (b V c V d) A (a V c V d) A (a V b)

A 3CNF formula is a CNF formula with exactly three literals per clause; the
previous example is not a 3CNF formula, since its first clause has four literals
and its last clause has only two. 3 Sat is just Sat restricted to 3CNF formulas:
Given a 3CNF formula, is there an assignment to the variables that makes the
formula evaluate to True?

We could prove that 3 Sat is NP-hard by a reduction from the more general
Sat problem, but it’s easier just to start over from scratch, by reducing directly
from CircuitSat.

Figure 12.5. A polynomial-time reduction from CircuitSat to 3SAT.

Given an arbitrary boolean circuit K, we transform K into an equivalent
3CNF formula in several stages. As we describe each stage, we also prove that
stage is correct.

• Make sure every and and or gate in K has exactly two inputs. If any gate has
k > 2 inputs, replace it with a binary tree of /< — 1 two-input gates. Call the
resulting circuit K'. The circuits K and K' are logically equivalent circuits,
so every satisfying input for K is a satisfying input for K' and vice versa.

• Transcribe K' into a boolean formula "fq with one clause per gate, exactly as
in our previous reduction to Sat. We already proved that every satisfying
input for <T>, can be transformed into a satisfying assignment for <T>, and vice
versa.

386

12.6. 3SAT (from Sat)

• Replace each clause in <!>] with a CNF formula. There are only three types of
clauses in one for each type of gate in K':

a = bAc < —» (a V b V c) A (a V b) A (a V c)
a = bVc <■—» (a V b V c) A (a V b) A (a V c)
a = b <—» (aVb)A(aVb)

Call the resulting CNF formula <f> 2 . Because <f> 1 and <f> 2 are logically equivalent
formulas, every satisfying assignment for "Jq is also a satisfying assignment
for <J> 2 , and vice versa. 7

• Replace each clause in $ 2 with a 3CNF formula. Every clause in <F 2 has at
most three literals. We can keep the three-literal clauses as-is. We expand
each two-literal clause into two three-literal clauses by introducing a new
variable. Finally, we expand any one-literal clause into four three-literal
clauses by introducing two new variables.

a V b < — * (a V b V x) A (a V b V x)

a < —» (a V x V y) A (a V x V y) A (a V x V y) A (a V x V y)

Call the final 3CNF formula <f> 3 . Every satisfying assignment for <F 2 can be
transformed into a satisfying assignment for <f> 3 by assigning arbitrary values
to the new variables (x and y). Conversely, every satisfying assignment for
<f> 3 can be transformed into a satisfying assignment for <l> 2 by ignoring the
new variables.

For example, our example circuit is transformed into the following 3CNF formula:

(36 V xT V xf) A (yf V x 1 V z 3 ) A (yf V x 1 V zf) A (yf V x 4 V z 2 ) A (yf V x 4 V zf)

A (y 2 v x 4 V Z 3 ) A (y 2 V x 4 V zf) A (yj V V z 4 ) A (j6 V x^ V zf)

A (y 3 v *3 V 36) A (yj V * 3 V z 5 ) A (yj V x 3 V zf) A (yj V y 2 V z 6 ) A (yj V y 2 V z^)

A (36 V yj V x 2 ) A (y 4 Vf)V z 7 ) A(y 4 VJ)Vi))A(y 4 Vy V z s ) A (y 4 V yf V T & )

A (y 5 V x 2 V z 9 ) A(y 5 Vx 2 Vz^)A(yVJ)V z 10 ) A (y( Vf)V zjj()

A (y 6 V x 5 V z n ) A (y 6 V x 5 V zfj") A (y( V x^ V z 12 ) A (y^ V x^ V z^)

A (36 V y 3 V y 5 ) A (y 7 V yj V z 13 ) A(y 7 V^V%)A(y 7 V^V z 14 ) A (y 7 V y^ V zff)

A (y 8 V y A V yj) A (36 V y 4 Vz 15 ) A (yj V y 4 V z^) A (y V y 7 V z 16 ) A 06 V y 7 v z^)

A (y 9 V y V ye) A (36 v y 8 v z 17 ) A 06 V y 6 V z 18 ) A (y V y 6 V z^) A (yj V y 8 V zff)

A (y 9 V z 19 V z 20 ) A(y,Vz^V z 20 ) A (y 9 V z 19 v %)) A (y 9 V z^ V z^)

7 This transformation from boolean circuit K to boolean formula $ 2 in conjunctive normal
form with at most three literals per clause was actually presented by Grigorii Tseitin in 1966,
five years before Levin announced his results about perebor problems; in the same paper, Tseitin
described the problem we now call CNF-Sat, perhaps for the first time.

387

12 . NP-Hardness

At first glance, this formula may look a lot more complicated than the original
circuit, but in fact, it’s only larger by a constant factor—each binary gate in
the original circuit has been transformed into at most five clauses. Even if the
formula size were a large polynomial function (like n 374 ) of the circuit size, we
would still have a valid reduction.

Our reduction transforms an arbitrary boolean circuit K into a 3CNF for¬
mula in polynomial time. Moreover, any satisfying input for the input
circuit K can be transformed into a satisfying assignment for <F 3 , and any
satisfying assignment for <f> 3 can be transformed into a satisfying input for K. In
other words, K is satisfiable if and only if <i> 3 is satisfiable. Thus, if 3 Sat can be
solved in polynomial time, then CircuitSat can be solved in polynomial time,
which implies that P = NP. We conclude 3 Sat is NP-hard.

12.7 Maximum Independent Set (from 3 Sat)

For the next few problems we consider, the input is a simple, unweighted,
undirected graph, and the problem asks for the size of the largest or smallest
subgraph satisfying some structural property.

Let G be an arbitrary graph. An independent set in G is a subset of the
vertices of G with no edges between them. The maximum independent set
problem, or simply MaxIndSet, asks for the size of the largest independent
set in a given graph. I will prove that MaxIndSet is NP-hard using a reduction
from 3Sat, as suggested by the following figure:

Given an arbitrary 3CNF formula <f>, we construct a graph G as follows.
Let k denote the number of clauses in <f>. The graph G contains exactly 3 k
vertices, one for each literal in <f>. Two vertices in G are connected by an
edge if and only if either (1) they correspond to literals in the same clause, or
(2) they correspond to a variable and its inverse. For example, the formula
(a V b V c) A (b V c V d) A (a V c V d) A (a V b V d) is transformed into the graph
shown in Figure 12.6.

Each independent set in G contains at most one vertex from each clause
triangle, because any two vertices in each triangle are connected. Thus, the
largest independent set in G has size at most k. I claim that G contains an

388

127 - Maximum Independent Set (from 3 Sat)

Figure 12.6. A graph derived from a satisfiable 3CNF formula with 4 clauses, and an independent set
of size 4.

independent set of size exactly k if and only if the original formula <f> is satisfiable.
As usual for “if and only if” statements, the proof consists of two parts.

=> Suppose $ is satisfiable. Fix an arbitrary satisfying assignment. By definition,
each clause in $ contains at least one True literal. Thus, we can choose
a subset S of k vertices in G that contains exactly one vertex per clause
triangle, such that the corresponding literals are all True. Because each
triangle contains at most one vertex in S, no two vertices in S are connected
by a triangle edge. Because every literal in S is True, no two vertices in S
are connected by a negation edge. We conclude that S is an independent
set of size kin G.

<f= On the other hand, suppose G contains an independent set S of size k. Each
vertex in S must lie in a different clause triangle. Suppose we assign the
value True to each literal in S; because contradictory literals are connected
by edges, this assignment is consistent. There may be variables x such that
neither x nor x corresponds to a vertex in S; we can set these variables to
any value we like. Because S contains one vertex in each clause triangle,
each clause in <J> contains (at least) one True literal. We conclude that <f> is
satisfiable.

Transforming the 3CNF formula <f> into the graph G takes polynomial time,
even if we do everything by brute force. Thus, if we could solve MaxIndSet
in polynomial time, then we could also solve 3 Sat in polynomial time, by
transforming the input formula <J> into a graph G and comparing the size of the
largest independent set in G with the number of clauses in <f>. But that would
imply P=NP, which is ridiculous! We conclude that MaxIndSet is NP-hard.

389

12 . NP-Hardness

12.8 The General Pattern

All NP-hardness proofs—and more generally, all polynomial-time reductions—
follow the same general pattern. To reduce problem X to problem Y in
polynomial time, we need to do three things:

1. Describe a polynomial-time algorithm to transform an arbitrary instance
of x of X into a special instance y of Y.

2. Prove that if x is a “good” instance of X, then y is a “good” instance of Y.

3. Prove that if y is a “good” instance of Y, then x is a “good” instance ofX.
(This is usually the part that causes the most trouble.)

Of course, developing a correct reduction doesn’t mean handling these three
tasks one at a time. First writing down an algorithm that seems to work and then
trying prove that it actually works is rarely successful, especially in time-limited
settings like exams. We must develop the algorithm, the “if” proof, and the
“only if” proof simultaneously.

To quote the late great Ricky Jay: 8 This is an acquired skill.

One point that confuses many students is that the reduction algorithm only
reduces “one way”—from X to Y —but the correctness proof needs to work
“both ways”. But the proofs are not actually symmetric. The “if” proof needs to
handle arbitrary instances of x, but the “only if” only needs to handle the special
instances of Y produced by the reduction algorithm. Exploiting this asymmetry
is actually the key to successfully designing correct reductions.

I find it useful to think in terms of transforming certificates —proofs that a
given instance is “good”—along with the instances themselves. For example,
a certificate for CircuitSat is a set of inputs that turns on the light bulb;
a certificate for Sat or 3Sat is a satisfying assignment; a certificate for MaxIndSet
is a large independent set. To reduce X to Y, we actually need to design three
algorithms, one for each of the following tasks:

• Transform an arbitrary instance x of X into a special instance y of Y in

polynomial time.

• Transform an arbitrary certificate for x into a certificate for y, and

• Transform an arbitrary certificate for y into a certificate for x.

The second and third tasks refer to the input and output of the first algorithm. The
certificate transformation needs to be reversible, not the instance transformation.
We never have to transform instances of Y, and we don’t need to think about
arbitrary instances of Y at all. Only the first algorithm needs to run in polynomial
time (although in practice, the second and third algorithms are almost always
simpler than the first).

8 from his 1996 off-Broadway show Ricky Jay and his 52 Assistants

390

12.g. Clique and Vertex Cover (from Independent Set)

For example, our reduction from CircuitSat to 3 Sat consists of three
algorithms:

• The first transforms an arbitrary boolean circuit K into a special 3CNF
boolean formula <f> 3 , in polynomial time. (Encode each wire as a variable
and each gate as a sub-formula, and then expand each sub-formula into
3CNF.)

• The second transforms an arbitrary satisfying input for K into a satisfying
assignment for <F 3 . (Trace the input through the circuit, transfer values from
each wire to the corresponding variable, and give any additional variables
arbitrary values.)

• The third transforms an arbitrary satisfying assignment for <f> 3 into a sat¬
isfying input for K. (Transfer values from each wire variable in $5 to the
corresponding wire in fC.)

The reduction works because it encodes any boolean circuit K into a highly
structured 3CNF formula <f> 3 . The specific structure of <F 3 restricts how it can be
satisfied; every satisfying assignment for <1> 3 must “come from” some satisfying
input for K. We don’t have to think about arbitrary 3CNF formulas at all.

Similarly, our reduction from 3Sat to MaxIndSet consists of three algorithms:

• The first transforms an arbitrary 3CNF formula <f> into a special graph G and
a specific integer k, in polynomial time.

• The second transforms an arbitrary satisfying assignment for <f> into an
independent set in G of size k.

• The third transforms an arbitrary independent set in G of size k into a
satisfying assignment for <f>.

Again, our first transformation encodes the input formula <J> into a highly
structured graph G and a specific integer k. The structure of G ensures that
every independent set of size k “comes from” a satisfying assignment for <F. We
don’t have to think about arbitrary graphs or arbitrary independent set sizes at
all.

12.9 Clique and Vertex Cover (from Independent Set)

A clique is another name for a complete graph, that is, a graph where every
pair of vertices is connected by an edge. The MaxClique problem asks for the
number of nodes in its largest complete subgraph in a given graph. A vertex
cover of a graph is a set of vertices that touches every edge in the graph. The
MinVertexCover problem asks for the size of the smallest vertex cover in a
given graph.

We can prove that MaxClique is NP-hard using the following easy reduction
from MaxIndSet. Any graph G has an edge-complement G with the same vertices,

391

12 . NP-Hardness

Figure 12.7. A graph whose largest independent set, largest clique, and smallest vertex cover all have
size 4.

but with exactly the opposite set of edges—uv is an edge in G if and only if uv
is not an edge in G. A set of vertices is independent in G if and only if the same
vertices define a clique in G. Thus, the largest independent in G has the same
vertices (and thus the same size) as the largest clique in the complement of G.

The proof that MinVertexCover is NP-hard is even simpler, because it relies
on the following easy observation: I is an independent set in a graph G = (V, E)
if and only if its complement V \ 7 is a vertex cover of the same graph G. Thus,
the largest independent set in any graph is just the complement of the smallest
vertex cover of the same graph! Thus, if the smallest vertex cover in an n-vertex
graph has size k, then the largest independent set has size n — k.

Figure 12.8. Easy reductions from MAXlNDSETto MAxCLiQUEand MinVertexCover.

392

12.1 o Graph Coloring (from 3 Sat)

A proper k-coloring of a graph G = (V, E) is a function C: V —» { 1 , 2 ,..., k}
that assigns one of k “colors” to each vertex, so that every edge has two different

12.10. Graph Coloring (from 3 Sat)

colors at its endpoints. (The “colors” are really just arbitrary labels, which for
simplicity we represent by small positive integers, rather than electromagnetic
frequencies, CMYK vectors, or Pantone numbers, for example.) The graph
coloring problem asks for the smallest possible number of colors in a legal
coloring of a given graph.

To prove that graph coloring is NP-hard, it suffices to consider the decision
problem 3C0LOR: Given a graph, does it have a proper 3 -coloring? We prove
3C0LOR is NP-hard using a reduction from 3 Sat. (Why 3 Sat? Because it has a 3
in it. You probably think I’m joking, but I’m not.) Given a 3CNF formula < 3 ?, we
construct a graph G that is 3-colorable if and only if <t> is satisfiable, as suggested
by the usual diagram.

3 SAT

True

True

3COLOR

0

transform
in 0 (n)
time

G

G is

3-colorable

0 is

satisfiable

3CNF

Boolean

formula

graph

False

False

G is not

O is not

3-colorable

satisfiable

We describe the reduction using a standard strategy of decomposing the
output graph G into gadgets, subgraphs that enforce various semantics of the
input formula <F. Decomposing reductions into separate gadgets is not only
helpful for understanding existing reductions and proving them correct, but for
designing new NP-hardness reductions. Our formula-to-graph reduction uses
three types of gadgets:

• There is a single truth gadget : a triangle with three vertices T, F, and X,
which intuitively stand for True, False, and Other. Since these vertices
are all connected, they must have different colors in any 3 -coloring. For the
sake of convenience, we will name those colors True, False, and Other.
Thus, when we say that a node is colored True, we mean that it has the
same color as vertex T.

• For each variable a, the graph contains a variable gadget, which is a triangle
joining two new nodes labeled a and a to node X in the truth gadget. Node a
must be colored either True or False, and therefore node a must be colored
either False or True, respectively.

Figure 12.g. The truth gadget and a variable gadget for a.

393

12 . NP-Hardness

• Finally, for each clause in <f>, the graph contains a clause gadget. Each clause
gadget joins three literal nodes (from the corresponding variable gadgets)
to node T (from the truth gadget) using five new unlabeled nodes and ten
edges, as shown below

In effect, each triangle in the clause gadget behaves like a “majority
gate”. In any valid 3-coloring, if the two vertices to the left of the triangle
have the same color, the rightmost vertex of the triangle must have the same
color; on the other hand, if the two left vertices have different colors, the
color of the right vertex can be chosen arbitrarily.

Figure 12.11. All valid 3-colorings of a "half gadget", up to permutations of the colors

It follows that there is no valid 3-coloring of a clause gadget where all
three literal nodes are colored False. On the other hand, any coloring of the
literal nodes with more than one color can be extended to a valid 3-coloring
of the clause gadget. The variable gadgets force each literal node to be
colored either True or False; thus, in any valid 3-coloring of the clause
gadget, at least one literal node is colored True.

The final graph G contains exactly one node T, exactly one node F, and exactly
two nodes a and a for each variable. For example, the formula (aVbVc)A
(b V c V d) A (a V c V d) A (a V b V d) that we previously used to illustrate the
MaxClique reduction would be transformed into the graph shown in Figure 12.12.
The 3-coloring is one of several that correspond to the satisfying assignment
a = c = True, b = d = False.

We’ve already done most of the work for a proof of correctness. If the formula
is satisfiable, then we can color the literal nodes according to any satisfying
assignment, and then (because each clause is satisfied) extend the coloring

394

12.11. Hamiltonian Cycle

Figure 12.12. The 3-colorable graph derived from a satisfiable 3CNF formula.

across every clause gadget. On the other hand, if the graph is 3-colorable, then
we can extract a satisfying assignment from any 3-coloring—at least one of the
three literal nodes in every clause gadget is colored True.

Because 3C0LOR is a special case of the more general graph coloring problem—
What is the minimum number of colors?—the more general optimization problem
is also NP-hard.

12.11 Hamiltonian Cycle

A Hamiltonian cycle in a graph is a cycle that visits every vertex exactly once.
(This is very different from an Eulerian cycle, which is actually a closed walk that
traverses every edge exactly once; Eulerian cycles are easy to find and construct
in linear time using depth-first search.) Here we consider two different proofs
that the Hamiltonian cycle problem for directed graphs is NP-hard.

From Vertex Cover

Our first NP-hardness proof reduces from the decision version of the vertex cover
problem. Given an undirected graph G and an integer k, we construct a directed
graph H, such that H has a Hamiltonian cycle if and only if G has a vertex cover
of size k. As in our previous reductions, the output graph H is composed of
several gadgets, each corresponding to certain features of the inputs G and k.

395

12 . NP-Hardness

• For each undirected edge uv in G, the directed graph H contains an edge
gadget consisting of four vertices (iz, v, in), (tz, v, out), (v, zz, in), (v, zz, out) and
six directed edges

(zz, v, in)^(zz, v, out) (iz, v, in)->(v, zz, in) (v, zz, in)-*(zz, v, in)

(v, zz, in)->(v, zz, out) (zz, v, out)->(v, zz, out) (v, tz, out)->(zz, v, out)

as shown in Figure 12.13. Each “in” vertex has an additional incoming edge,
and each “out” vertex has an additional outgoing edge. Any Hamiltonian
cycle in H must pass through an edge gadget in one of three ways—either
straight through on both sides, or with a detour from one side to the other
and back. Eventually, these options will correspond to both zz and v, only zz,
or only v belonging to some vertex cover.

Figure 12.13. An edge gadget and its only possible intersections with a Hamiltonian cycle.

• For each vertex tz in G, all the edge gadgets for incident edges uv are
connected in H into a single directed path, which we call a vertex chain.
Specifically, suppose vertex zz has d neighbors v 1 ,v 2 , ■ ■ ■ ,v d . Then H has
d — 1 additional edges (zz, v ; , out)-z(zz, v i+1 , in) for each z from 1 to d — 1.

• Finally, H also contains k cover vertices x 1 ,x 2 , ■ ■ ., x k . Each cover vertex has
a directed edge to the first vertex in each vertex chain, and a directed edge
from the last vertex in each vertex chain.

An example of our complete transformation is shown in Figure 12.14. As usual,
we prove that the reduction is correct in two stages.

=> First, suppose C = {u 1 ,u 2 ,... ,u k } is a vertex cover of G. Then we can
construct a Hamiltonian cycle in H as follows. For each index i from 1 to k,

396

12.11. Hamiltonian Cycle

Figure 12.14. Reduction from VertexCover to HamiltonianCycle.

we traverse a path from cover vertex x n through the vertex chain for u n to
cover vertex x i+1 (or cover vertex x 1 ii i = k). As we traverse the chain for
each vertex u h we determine how to proceed from each node (u h v, in) as
follows:

- If v G C, just follow the edge (zz ; , v, in)->(u ; , v, out).

- If v ^ C, detour through (u ; , v, in)->(v, tz ; , in)^(v, u ; , out)->(u ; , v, out).

Thus, for each edge uv of G, the Hamiltonian cycle visits (iz, v, in) and
(tz, v, out) as part of zz’s vertex chain if iz € C and as part of v’s vertex chain
otherwise. See Figure 12.15.

k=2

Figure 12.15. Every vertex cover of size fc in G corresponds to a Hamiltonian cycle inH and vice versa.

397

12 . NP-Hardness

<= On the other hand, suppose H contains a Hamiltonian cycle C. This
cycle must contain an edge from each cover vertex to the start of some
vertex chain. Our case analysis of edge gadgets inductively implies that
after C enters the vertex chain for some vertex iz, it must traverse the en¬
tire vertex chain. Specifically, at each vertex (iz, v, in), the cycle must
contain either the single edge (iz, v, in)->(zz, v, out) or the detour path
(iz, v, in)^(v, u, in)->(v, iz, out)^(iz, v, out), followed by an edge to the next
edge gadget in tz’s vertex chain, or to a cover vertex if this is the last such
edge gadget. In particular, if C contains the detour edge (iz, v, in)^(v, iz, in),
it does not contain edges between any cover vertex and v’s vertex chain.
It follows that C traverses exactly k vertex chains. Moreover, these vertex
chains describe a vertex cover of the original graph G, because C visits the
vertex (iz, v, in) for every edge tzv in G.

We conclude that G has a vertex cover of size k if and only if H contains a
Hamiltonian cycle. The transformation from G to H takes at most 0 (zi 2 ) time;
it follows that the directed Hamiltonian cycle problem is NP-hard.

From 3SAT

We can also prove that the Hamiltonian cycle problem is NP-hard by reducing
directly from 3 Sat. Given a 3CNF formula <f> with n variables x 1 ,x 2 , ■ ■ ■ ,x n
and k clauses ci,c 2 ,...,c k , we construct a directed graph H that contains a
Hamiltonian cycle if and only if <f> is satisfiable, as follows.

For each variable x h we construct a variable gadget, which consists of a
doubly-linked list of 2 k vertices (z, 0), (z, 1),..., (z, 2 k), connected by edges
(z, j — l)->(z,j) and (z, j)->(z, j — 1 ) for each index j. We connect the first and
last nodes in each adjacent pair of variable gadgets by adding edges

(z, 0)—>(i + 1,0) (i,2fc)-Ki + l,0) (z, 0)->(z + 1,2k) (i,2fc)->(i + 1,2k)

for each index i; we also connect the endpoints of the first and last variable
gadgets with the edges

(n, 0 )-»(l, 0 ) (n, 2 IcHl, 0 ) (n, 0 Wl, 2 fc) (n, 2 fc)->(l, 2 fc).

The resulting graph G has exactly 2 n Hamiltonian cycles, one for each assignment
of boolean values to the n variables of <F. Specifically, for each z, we traverse the
zth variable gadget from left to right if x l = True and right to left if x l = False.
See Figure 12.16.

Now we extend G to a larger graph H by adding a clause vertex [j] for each
clause Cj, connected to the variable gadgets by six edges, as shown in Figure 12.17.
For each positive literal x t in Cj, we add the edges (z, 2 j — 1 )—»•[_/]—»-(i, 2 j), and
for each negative literal x t in Cy, we add the edges (z, 2 _/)—>[j]—>-(i, 2 j — 1 ). The

398

12.11. Hamiltonian Cycle

Figure 12.16. Left: Variable gadgets in G. Right: The Hamiltonian cycle in G for the assignment a =
b = d = True and c = False

(a V b V c) A (b V c V d) A (a V c V d) A (a V b V d)

(a V b V c) A (b V c V d) A (a V c V d) A (a V b V d)

Figure 12.17. Left: Clause gadgets in H. Right: A Hamiltonian cycle in H corresponding to the satisfying
assignment a = b = d = True and c = False.

connections to the clause vertices guarantee that a Hamiltonian cycle in G
can be extended to a Hamiltonian cycle in H if and only if the corresponding
variable assignment satisfies <f>. Exhaustive case analysis now implies that H has
a Hamiltonian cycle if and only if $ is satisfiable.

Transforming the formula $ into the graph H takes 0 (kn) time, which is at
most quadratic in the total length of the formula; we conclude that the directed
Hamiltonian cycle problem is NP-hard.

Variants and Extensions

Trivial modifications of the previous reductions imply that the Hamiltonian
path problem is also NP-hard. A Hamiltonian path in a graph G is of course a
simple path that visits every vertex of G exactly once. In fact, there are simple
polynomial-time reductions from HamiltonianCycle to HamiltonianPath and
vice versa. I’ll leave the details of these reductions as easy exercises.

399

12 . NP-Hardness

Both of the previous reductions deal with directed graphs, but the corre¬
sponding question in undirected graph is also NP-hard. In fact, there is a
relatively simple reduction from the directed Hamiltonian cycle problem to
the undirected Hamiltonian cycle problem; again, I’ll leave the details of this
reduction as an entertaining exercise.

Finally, the infamous traveling salesman problem asks to find the shortest
Hamiltonian cycle (or path) in a graph with weighted edges. Since finding the
shortest cycle/path is obviously harder than determining if a cycle/path exists at
all—Consider a graph where every edge has weight 1 !—the traveling salesman
problem is also NP-hard.

12.12 Subset Sum (from Vertex Cover)

The next problem that we prove NP-hard is the SubsetSum problem considered
in Chapter 2: Given a set X of positive integers and an integer T, determine
whether X has a subset whose elements sum to T.

We once again reduce from VertexCover. Given a graph G and an integer k,
we need to compute a set X of positive integers and an integer T, such that X
has a subset that sums to T if and only if G has an vertex cover of size k. Our
transformation uses just two “gadgets”, which are integers representing the
vertices and edges in G.

Number the edges of G arbitrarily from 0 to m — 1 . Our set X contains the
integer b, := 4 ! for each edge i, and the integer

a v := 4 m + Xi 4 ‘

is A(v)

for each vertex v, where A(v) is the set of edges that have v has an endpoint.
Alternately, we can think of each integer in A as an (m + l)-digit number written
in base 4 . The mth digit is 1 if the integer represents a vertex, and 0 otherwise;
and for each i < m, the ith digit is 1 if the integer represents edge i or one of
its endpoints, and 0 otherwise. Finally, we set the target sum

m—1

T:=k- 4 m + J] 2 ' 4 ! ’.

i =0

Now let’s prove that the reduction is correct.

=> First, suppose there is a vertex cover of size k in the original graph G.
Consider the subset X c c X that includes a v for every vertex v in the vertex
cover, and b f for every edge i that has exactly one vertex in the cover. The
sum of these integers, written in base 4, has a 2 in each of the first m digits;
in the most significant digit, we are summing exactly k l’s. Thus, the sum
of the elements of X c is exactly T.

400

12.12. Subset Sum (from Vertex Cover)

<= On the other hand, suppose there is a subset X' c X that sums to t.
Specifically, we must have

2> v +2> =t

veV' ie£'

for some subsets V 1 c y and E 7 c Again, if we sum these base-4 numbers,
there are no carries in the first m digits, because for each i there are only
three numbers inX whose i th digit is 1 . Each edge number b t contributes
only one 1 to the ith digit of the sum, but the ith digit of t is 2 . Thus, for
each edge in G, at least one of its endpoints must be in V 1 . In other words,
V is a vertex cover. On the other hand, only vertex numbers are larger than
4 m , and [T/ 4 m J = k, so V' has at most k elements. (In fact, it’s not hard to
see that V' has exactly k elements.)

For example, given the four-vertex graph G = (V,E) where V = {u, v,w,x) and
E = {uv, uw, vw, vx, wx), our set X might contain the following base-4 integers:

a u := 111000 4 =

1344

b uv

= 010000 4 =

256

a v := 110110 4 =

1300

buw

= 001000 4 =

64

a w := 101101 4 =

1105

b vw

= 000100 4 =

16

a x := 100011 4 =

1029

b vx

= 000010 4 =

4

bwx

= 000001 4 =

1

If we are looking for a vertex cover of size 2 , our target sum would be T :=
222222 4 = 2730 . Indeed, the vertex cover {v, w) corresponds to the subset
{a v ,a w , b uv , b uw , b vx , b wx }, whose sum is 1300 + 1105 + 256 + 64+4 + 1 = 2730 .

The reduction can clearly be performed in polynomial time. We’ve already
proved that VertexCover is NP-hard, so it follows that SubsetSum is NP-hard.

Caveat Reductor!

One subtle point must be emphasized here. 3 oo-something pages ago, back in
Chapter 3, we developed a dynamic programming algorithm to solve SubsetSum
in O(nT) time. Isn’t this a polynomial-time algorithm? Didn’t we just prove that
P=NP? Hey, where’s my million dollars?

Alas, life is not so simple. True, the running time is a polynomial function
of n and T, but in order to qualify as a true polynomial-time algorithm, the
running time must be a polynomial function of the number of bits required to
represent the input. The values of the elements of A and the target sum T could
be exponentially larger than the number of input bits. Indeed, the reduction
we just described produces a value of T that is exponentially larger than the
size of our original input graph, which would force our dynamic programming
algorithm to run in exponential time.

401

12 . NP-Hardness

Algorithms like this are said to run in pseudo-polynomial time, and any
NP-hard problem with such an algorithm is called weakly NP-hard. Equivalently,
a weakly NP-hard problem is one that can be solved in polynomial time when
all input numbers are represented in unary (as a sum of Is), but becomes
NP-hard when all input numbers are represented in binary. If a problem is
NP-hard even when all the input numbers are represented in unary, we say
that the problem is strongly NP-hard. A good example of a strongly NP-hard
problem is TravelingSalesman, which remains NP-hard even if the input graph
is complete and all edge weights are equal to 1 or 2.

12.13 Other Useful NP-hard Problems

Literally thousands of problems have been proved to be NP-hard. Here I will list
a few NP-hard problems that are useful in deriving reductions. I won’t describe
the NP-hardness proofs for these problems in detail, but you can find most of
them in Garey and Johnson’s classic Scary Black Book of NP-Completeness. 9 All
of the problems I’ve discussed so far, and most of the problems in the following
list, were proved NP-hard in a single landmark 1972 paper by Richard Karp. 10

• PlanarCircuitSat: Given a boolean circuit that can be embedded in the
plane so that no two wires cross, is there an input that makes the circuit
output True? This problem can be proved NP-hard by reduction from the
general circuit satisfiability problem, by replacing each crossing with a small
assemblage of gates.

• 1-IN-3SAT: Given a 3CNF formula, is there an assignment of values to the
variables so that each clause contains exactly one True literal? This problem
can be proved NP-hard by reduction from the usual 3Sat.

• NotAllEqual3Sat: Given a 3CNF formula, is there an assignment of
values to the variables so that every clause contains at least one True literal
and at least one False literal? This problem can be proved NP-hard by
reduction from the usual 3 Sat.

• Planar3Sat: Given a 3CNF boolean formula, consider a bipartite graph
whose vertices are the clauses and variables, where an edge indicates that
a variable (or its negation) appears in a clause. If this graph is planar, the
3CNF formula is also called planar. The Planar3Sat problem asks, given a
planar 3CNF formula, whether it has a satisfying assignment. This problem
can be proved NP-hard by reduction from PlanarCircuitSat. 11

9 Michael Garey and David Johnson. Computers and Intractability: A Guide to the Theory of
NP-Completeness. W. H. Freeman and Co., 1979. Yes, it really is black.

“Later performed off-Broadway as Richard Karp and his 21 Assistants, for which Karp won a
well-deserved Tony Turing award.

“Surprisingly, PlanarNotAllEqual3Sat is solvable in polynomial time!

402

12.13. Other Useful NP-hard Problems

• Exact3DimensionalMatching or X3M: Given a set S and a collection of
three-element subsets of S, called triples, is there a sub-collection of disjoint
triples that exactly cover S? This problem can be proved NP-hard by a
reduction from 3Sat.

• Partition: Given a set S of n integers, are there subsets A and B such that
AUB = S,ADB — 0 , and

2 >= 2 >

aeA beB

This problem can be proved NP-hard by a simple reduction from SubsetSum.
Like SubsetSum, the Partition problem is only weakly NP-hard.

• 3Partition: Given a set S of 3 n integers, can it be partitioned into n disjoint
three-element subsets, such that every subset has exactly the same sum?
Despite the similar names, this problem is very different from Partition;
sorry, I didn’t make up the names. This problem can be proved NP-hard by
reduction from X3M. Unlike Partition, the 3Partition problem is strongly
NP-hard; it remains NP-hard even if every input number is at most n 3 .

• SetCover: Given a collection of sets S = {S 1; S 2 , •.., S m }, find the smallest
sub-collection of S ; ’s that contains all the elements of S ; . This problem is
a generalization of both VertexCover and X3M.

• HittingSet: Given a collection of sets S = {S 1; S 2 ,... ,S m }, find the mini¬
mum number of elements of [J ( S ; that hit every set in S. This problem is
also a generalization of VertexCover.

• LongestPath: Given a non-negatively weighted graph G and two vertices u
and v, what is the longest simple path from u to v in the graph? A path is
simple if it visits each vertex at most once. This problem is a generalization
of the HamiltonianPath problem. Of course, the corresponding shortest
path problem can be solved in polynomial time.

• SteinerTree: Given a weighted, undirected graph G with some of the
vertices marked, what is the minimum-weight subtree of G that contains
every marked vertex? If every vertex is marked, the minimum Steiner tree
is just the minimum spanning tree; if exactly two vertices are marked, the
minimum Steiner tree is just the shortest path between them. This problem
can be proved NP-hard by reduction from VertexCover.

• Max2Sat: Given a Boolean formula in conjunctive normal form, with exactly
two literals per clause, find a variable assignment that maximizes the number
of clauses with at least one True literal. This problem can be proved NP-hard
by reduction from 3Sat. The simpler decision problem 2Sat, which asks if
there is an assignment that satisfies every clause, can actually be solved in
polynomial time.

403

12 . NP-Hardness

• MaxCut: Given an undirected graph G = (V,E), find a subset S CV that
maximizes the number of edges with exactly one endpoint in S. Equivalently,
find the largest bipartite subgraph of G. This problem can be proved NP-hard
by reduction from Max2Sat.

In addition to these dry but useful problems, most interesting puzzles and soli¬
taire games have been shown to be NP-hard, or to have NP-hard generalizations.
(Arguably, if a game or puzzle isn’t at least NP-hard, it isn’t interesting!) Here
are some examples you may find familiar:

• Minesweeper (from CircuitSat) 12

• Sudoku (utlimately from 3Sat) 13

• Tetris (from 3Partition) 14

• Klondike, aka “Solitaire” (from 3Sat) 15

• Pac-Man (from HamiltonianCycle) 16

• Super Mario Brothers (from 3Sat) 17

• Candy Crush Saga (from a variant of 3 Sat) 18

• Threes/2048 (from 3SAT) 19

• Trainyard (from DominatingSet) 20

“Richard Kaye. Minesweeper is NP-complete. Mathematical Intelligencer 22(2)19-15, 2000.
But see also: Allan Scott, Ulrike Stege, and Iris van Rooij. Minesweeper may not be NP-complete
but is hard nonetheless. Mathematical Intelligencer 33(4):s-i7, 2011.

13 Takayuki Yato and Takahiro Seta. Complexity and completeness of finding another solution
and its application to puzzles. IEICE Transactions on Fundamentals of Electronics, Communications
and Computer Sciences E86-A(s) 11052-1060, 2003. http://www-imai.is.S-U-tokyo.ac.jp/~yato/
data2/MasterThesis.pdf.

14 Ron Breukelaar*, Erik D. Demaine, Susan Hohenberger*, Hendrik J. Hoogeboom, Walter A.
Rosters, and David Liben-Nowell*. Tetris is hard, even to approximate. International Journal of
Computational Geometry and Applications 14:41-68, 2004.

15 Luc Longpre and Pierre McKenzie. The complexity of Solitaire. Proceedings of the 32nd
International Mathematical Foundations of Computer Science, 182-193, 2007.

l6 Giovanni Viglietta. Gaming is a hard job, but someone has to do it! Theory of Computing
Systems, 54(41:595-621, 2014. http://giovanniviglietta.com/papers/gaming2.pdf.

17 Greg Aloupis, Erik D. Demaine, Alan Guo, and Giovanni Viglietta. Classic Nintendo
games Are (computationally) hard. Theoretical Computer Science 586:135-160, 2015. http:
//arxiv.org/abs/1203.1895.

l8 Luciano Guala, Stefano Leucci, Emanuele Natale. Bejeweled, Candy Crush and other
match-three games are (NP-)hard. Proc. 2014 IEEE Conference on Computational Intelligence and
Games, 2014. http://arxiv.0rg/abs/1403.5830.

19 Further evidence for the “rule of three”. Stefan Langerman and Yushi Uno. Threes!, Fives,
1024!, and 2048 are Hard. Proc. 8th International Conference on Fun with Algorithms, 2016.
https://arxiv.0rg/abs/1505.04274.

2 °Matteo Almanza, Stefano Leucci, and Alessandro Panconesi. Trainyard is NP-Hard. Proc.
8th International Conference on Fun with Algorithms, 2016. https://arxiv.0rg/abs/1603.00928.

404

12 . 14 - Choosing the Right Problem

• Shortest n x n x n Rubik’s cube solution (from HamiltonianCycle ). 21

This list is necessarily incomplete, thanks to a limited footnote budget. 22 As of
December 2018, nobody has published a proof that a generalization of Cookie
Clicker or Ultimate Paperclips is NP-hard, but I’m sure it’s only a matter of time.

12.14 Choosing the Right Problem

One of the most difficult steps in proving that a problem is NP-hard is choosing
the best problem to reduce from. The Cook-Levin Theorem implies that if there
is a reduction from any NP-hard problem to problem X, then there is a reduction
from every NP-complete problem to problem X, but some problems work better
than others. There’s no systematic method for choosing the right problem, but
here are a few useful rules of thumb.

• If the problem asks how to assign bits to objects, or to choose a subset of
objects, or to partition objects into two different subsets, try reducing from
some version of Sat or Partition.

• If the problem asks how to assign labels to objects from a small fixed set,
or to partition objects into a small number of subsets, try reducing from
k Color or even 3 Color.

• If the problem asks to arrange a set of objects in a particular order, try reduc¬
ing from HamiltonianCycle or HamiltonianPath or TravelingSalesman.

• If the problem asks to find a small subset satisfying some constraints, try
reducing from MinVertexCover.

• If the problem asks to find a large subset satisfying some constraints, try
reducing from MaxIndependentSet or MaxClique or Max2Sat.

• If the problem asks to partition objects into a large number of small subsets,
try reducing from 3Partition.

• If the number 3 appears naturally in the problem, try 3 Sat or 3C0LOR or
X3M or 3Partition. (No, this is not a joke.)

• If all else fails, try 3 Sat or even CircuitSat!

I do not recommend trying to reduce from Tetris, SuperMarioBros, or
Trainyard. You really want to choose a starting problem that is as simple as
possible, while still capturing some feature of your problem that makes it difficult
to solve.

21 Erik D. Demaine, Sarah Eisenstat, and Mikhail Rudoy. Solving the Rubik’s Cube optimally
is NP-complete. Proc. 35th Symposium on Theoretical Aspects of Computer Science, 2018. https:
//arxiv.org/abs/1706.06708.

22 See https://xkcd.com/1208/

405

12. NP-Hardness

12.15 A Frivolous Real-World Example

Draughts is a family of board games that have been played for thousands of
years. Most Americans are familiar with the version called checkers or English
draughts, but the most common variant worldwide, known as international
draughts or Polish draughts, originated in the Netherlands in the 16th century.
For a complete set of rules, the reader should consult Wikipedia; here a few
important differences from the Anglo-American game:

• Flying kings: As in checkers, a piece that ends a move in the row closest to
the opponent becomes a king and gains the ability to move backward. Unlike
in checkers, however, a king in international draughts can move any distance
along a diagonal line in a single turn, as long as the intermediate squares
are empty or contain exactly one opposing piece (which is captured).

• Forced maximum capture: In each turn, the moving player must capture
as many opposing pieces as possible. This is distinct from the forced-capture
rule in checkers, which requires only that each player must capture if possible,
and that a capturing move ends only when the moving piece cannot capture
further. In other words, checkers requires capturing a locally maximal set of
opposing pieces on each turn; whereas, international draughts requires a
globally maximum capture.

• Capture subtleties: As in checkers, captured pieces are removed from the
board only at the end of the turn. Any piece can be captured at most once.
Thus, when an opposing piece is jumped, that piece remains on the board
but cannot be jumped again until the end of the turn.

For example, in the first position shown below, each circle represents a piece,
and doubled circles represent kings. Black must make the first indicated move,
capturing five white pieces, because it is not possible to capture more than five
pieces, and there is no other move that captures five. Black cannot extend
his capture further, either northeast or southeast, because the captured White
pieces remain on the board until his turn is over. Then White must make the
second indicated move, thereby winning the game.

Figure 12.18. Two forced(!) moves in international draughts; doubled circles are kings.

406

The actual game, which is played on a 10 x 10 board with 20 pieces of
each color, is computationally trivial; we can precompute the optimal move for

12 . 15 - A Frivolous Real-World Example

both players in every possible board configuration and hard-code the results
into a lookup table of constant size. Sure, it’s a big constant, but it’s still just a
constant!

But consider the natural generalization of international draughts to an n x n
board. In this setting, finding a legal move is actually NP-hard! The following
reduction from the Hamiltonian cycle problem in directed graphs was discovered
by Bob Hearn in 2010. 23 In most two-player games, finding the best move is
NP-hard (or worse). This is the only example I know of a game—and moreover
a real game played by millions of people for centuries—where just following the
rules is NP-hard!

Given an undirected graph G with n vertices, we construct a board con¬
figuration for international draughts, such that White can capture a certain
number of black pieces in a single move if and only if G has a Hamiltonian
cycle. We treat G as a directed graph, with two arcs u->v and v-m in place of
each undirected edge uv. Number the vertices arbitrarily from 1 to n. The final
draughts configuration has several gadgets:

• The vertices of G are represented by rabbit-shaped vertex gadgets, which are
evenly spaced along a horizontal line. Each arc i->j is represented by a path
of two diagonal line segments from the “right ear” of vertex gadget i to the
“left ear” of vertex gadget j. The path for arc i-*j is located above the vertex
gadgets if i < j, and below the vertex gadgets if i > j.

Figure 12.19. A high-level overview of the reduction from Hamiltonian cycle to international draughts.

• The bulk of each vertex gadget is a diamond-shaped region called a vault.
The wall s of the vault are composed of two solid layers of black pieces,
which cannot be captured; these pieces are drawn as gray circles in the
figures. There are N capturable black pieces inside each vault, for some
large integer N to be determined later. A white king can enter the vault

23 See Theoretical Computer Science Stack Exchange: http://cstheory.stackexchange.eom/a/
1999/111.

407

12. NP-Hardness

through the “right ear”, capture every internal piece, and then exit through
the “left ear”. Both ears are hallways, again with walls two pieces thick,
with gaps where the arc paths end to allow the white king to enter and
leave. The lengths of the “ears” can easily be adjusted to align with the
other gadgets.

Figure 12.20. Left: A vertex gadget with three entrances and three exits. Right: A white king emptying
the vault. Gray circles are black pieces that cannot be captured.

• For each arc i-*j, we have a corner gadget, which allows a white king leaving
vertex gadget i to be redirected to vertex gadget j.

• Finally, wherever two arc paths cross, we have a crossing gadget; these
gadgets allow the white king to traverse either arc path, but forbid switching
from one arc path to the other.

Figure 12.21. Left: One of two paths through a corner gadget. Right: One of two paths through a
crossing gadget.

A single white king starts at the bottom corner of one of the vaults. In any
legal move, this king must alternate between traversing entire arc paths and
clearing vaults. The king can traverse the various gadgets backward, entering
each vault through the exit and vice versa. But the reversal of a Hamiltonian
cycle in G is another Hamiltonian cycle in G, so walking backward is fine.

If there is a Hamiltonian cycle in G, the white king can capture at least nN
black pieces by visiting each of the other vaults and returning to the starting
vault. On the other hand, if there is no Hamiltonian cycle in G, the white king
can capture at most half of the pieces in the starting vault, and thus can capture

408

2.16. On Beyond Zebra

at most (n — 1/2 )N + 0(n 3 ) enemy pieces altogether. The 0(n 3 ) term accounts
for the corner and crossing gadgets; each edge passes through one corner gadget
and at most n 2 /2 crossing gadgets.

To complete the reduction, we set N = n 4 . Summing up, we obtain an
0(n 5 ) x 0(n 5 ) board configuration, with 0(n 5 ) black pieces and one white king.
We can clearly construct this board configuration by brute force in polynomial
time. Figure 12.22 shows a complete example of the construction.

Figure 12.22. The final draughts configuration for the 4-vertex graph in Figure 12.19. (The green arrows
are not actually part of the configuration.)

It is still open whether the following related question is NP-hard: Given an
n x n board configuration for international draughts, can (and therefore must )
White capture all the black pieces in a single turn?

*1 2.16 On Beyond Zebra

P and NP are only the first two steps in an enormous hierarchy of complexity
classes. To close this chapter (and the book), let me describe a few more classes
of interest.

409

12. NP-Hardness

Polynomial Space

PSPACE is the set of decision problems that can be solved using polynomial
space. Every problem in NP (and therefore in P) is also in PSPACE. It is generally
believed that NP f PSPACE, but nobody can even prove that P / PSPACE.
A problem n is PSPACE-hard if, for any problem fF that can be solved using
polynomial space, there is a polynomial-time many-one reduction from rF to n.
If any PSPACE-hard problem is in NP, then PSPACE=NP; similarly, if any
PSPACE-hard problem is in P, then PSPACE=P.

The canonical PSPACE-hard problem is the quantified boolean formula
problem, or QBF : Given a boolean formula $ that may include any number
of universal or existential quantifiers, but no free variables, is $ equivalent to
True? For example, the following expression is a valid input to QBF:

3a: Vb: 3c: (Vd :aVbVcVd)<=> ((b A c) V (3e: (a => e) V (c 7 ^ a A e))).

Sat is equivalent to the special case of QBF where the input formula contains
only existential quantifiers. QBF remains PSPACE-hard even when the input
formula must have all its quantifiers at the beginning, the quantifiers strictly
alternate between 3 and V, and the quantified proposition is in conjunctive
normal form, with exactly three literals in each clause, for example:

3a: Vb: 3c: Vd: ((a VbVc)A(bVcVd)A(aVcVd)A(aVbV d))

This restricted version of QBF can also be phrased as a two-player strategy
question. Suppose two players, Alice and Bob, are given a 3 CNF predicate
with free variables x 1 , x 2 , ..., x n . The players alternately assign values to the
variables in order by index—Alice assigns a value to x 1 , Bob assigns a value
to x 2 , and so on. Alice eventually assigns values to every variable with an
odd index, and Bob eventually assigns values to every variable with an even
index. Alice wants to make the expression True, and Bob wants to make
it False. Assuming Alice and Bob play perfectly, who wins this game? Not
surprisingly, most two-player games 24 like tic-tac-toe, reversi, checkers, go,
chess, and mancala—or more accurately, appropriate generalizations of these
constant-size games to arbitrary board sizes—are PSPACE-hard.

Another canonical PSPACE-hard problem is NFA totality: Given a non-
deterministic finite-state automaton M over some alphabet E, does M accept
every string in S*? The closely related problems NFA equivalence (Do two given
NFAs accept the same language?) and NFA minimization (Find the smallest NFA
that accepts the same language as a given NFA) are also PSPACE-hard, as are

24 For a good (but inevitably aging) overview of known results on the computational complexity
of games and puzzles, see Erik Demaine and Bob Hearn’s monograph Games, Puzzles, and
Computation (CRC Press, 2009).

410

2.16. On Beyond Zebra

the corresponding questions about regular expressions. (The corresponding
questions about deterministic finite-state automata are actually solvable in
polynomial time.)

Exponential Time

The next significantly larger complexity class, EXP (also called EXPTIME), is
the set of decision problems that can be solved in exponential time, that is,
using at most 2 n steps for some constant c > 0. Every problem in PSPACE (and
therefore in NP (and therefore in P)) is also in EXP. It is generally believed
that PSPACE £ EXP, but nobody can even prove that NP EXP. A problem n is
EXP-hard if, for any problem II' that can be solved in exponential time, there is
a polynomial-time many-one reduction from n / to IT If any EXP-hard problem is
in PSPACE, then EXP=PSPACE; similarly, if any EXP-hard problem is in NP, then
EXP=NP. We do know that P EXP; in particular, no EXP-hard problem is in P.

Natural generalizations of many interesting 2 -player games—like checkers,
chess, mancala, and go—are actually EXP-hard. The boundary between PSPACE-
hard games and EXP-hard games is rather subtle. For example, there are three
ways to draw in chess (the standard 8x8 game): stalemate (the player to move
is not in check but has no legal moves), repeating the same board position three
times, or moving fifty times without capturing a piece. The n x n generalization
of chess is either in PSPACE or EXP-hard depending on how we generalize these
rules. If we declare a draw after (say) n 3 capture-free moves, then every game
must end after a polynomial number of moves, so we can simulate all possible
games from any given position using only polynomial space. On the other hand,
if we ignore the capture-free move rule entirely, the resulting game can last an
exponential number of moves, so there no obvious way to detect a repeating
position using only polynomial space; indeed, this version of n x n chess is
EXP-hard.

Excelsior!

Naturally, even exponential time is not the end of the story. NEXP is the class
of decision problems that can be solve in nondeterministic exponential time;
equivalently, a decision problem is in NEXP if and only if, for every Yes instance,
there is a proof of this fact that can be checked in exponential time. EXPSPACE
is the set of decision problems that can be solved using exponential space. Even
these larger complexity classes have hard problems; for example, if we add the
intersection operator n to the syntax of regular expressions, deciding whether
two such expressions describe the same language is EXPSPACE-hard. Beyond
EXPSPACE are complexity classes with doubly-ex ponential resource bounds

411

12. NP-Hardness

(EEXP, NEEXP, and EEXPSPACE), then triply exponential resource bounds
(EEEXP, NEEEXP, and EEEXPSPACE), and so on ad infinitum.

All these complexity classes can be ordered by inclusion:

p c NP c PSPACE C EXP C NEXP C EXPSPACE C EEXP C NEEXP C • • •

Most complexity theorists strongly believe that every inclusion in this sequence
is strict; that is, no two of these complexity classes are equal. However, the
strongest result that has been proved is that every class in this sequence is strictly
contained in the class three steps later in the sequence. For example, we have
proofs that P / EXP and PSPACE ^ EXPSPACE, but not whether P / PSPACE or
NP 7^ EXP.

The limit of this series of increasingly exponential complexity classes is the
class ELEMENTARY of decision problems that can be solved using time or space
bounded by a function the form 2 n for some constant integer k, where

2t k n~{ n _ iffc = °’

[2 2 ^ ' n otherwise.

For example, 2 j 1 n = 2 n and 2 } 2 n = 2 2 ".

It may be tempting to conjecture that every natural decidable problem can
be solved in elementary time, but in fact this conjecture is incorrect. Consider
the extended regular expressions defined by recursively combining (possibly
empty) strings over some finite alphabet by concatenation (xy), union (x + y),
Kleene closure (x*), and negation (x). For example, the extended regular
expression (0 + 1 )*00(0 + 1 )* represents the set of strings in {0, 1 }* that do not
contain two 0 s in a row. It is possible to determine algorithmically whether
two extended regular expressions describe identical languages, by recursively
converting each expression into an equivalent NFA, converting each NFA into a
DFA, and then minimizing the DFA. However, the running time of this algorithm
has the non-elementary bound 2 '[' e ('d 2, intuitively because each layer of
recursive negation can increase the number of states exponentially. In fact,
Larry Stockmeyer proved in 1974 that this problem cannot be solved in merely
elementary time, even if we forbid Kleene closure!

Exercises

1. (a) Describe and analyze and algorithm to solve Partition in time O(nM),
where n is the size of the input set and M is the sum of the absolute
values of its elements.

(b) Why doesn’t this algorithm imply that P=NP?

412

Exercises

2. Consider the following problem, called BoxDepth: Given a set of n axis-
aligned rectangles in the plane, how big is the largest subset of these
rectangles that contain a common point?

(a) Describe a polynomial-time reduction from BoxDepth to MaxClique.

(b) Describe and analyze a polynomial-time algorithm for BoxDepth. [Hint:
Off') time should be easy, but 0 {n log n) time is possible.]

(c) Why don’t these two results imply that P=NP?

3. A boolean formula is in disjunctive normal form (or DNFj if it consists of a
disjunction (Or) or several terms, each of which is the conjunction (And) of
one or more literals. For example, the formula

(x A y A zj V [y A z) V (x A y A z)

is in disjunctive normal form. DNF-Sat asks, given a boolean formula in
disjunctive normal form, whether that formula is satisfiable.

(a) Describe a polynomial-time algorithm to solve DNF-Sat.

(b) What is the error in the following argument that P=NP?

Suppose we are given a boolean formula in conjunctive normal form with
at most three literals per clause, and we want to know if it is satisfiable.

We can use the distributive law to construct an equivalent formula in dis¬
junctive normal form. For example,

(xVyVz)A(xVy) <=> (x A y) V (y A x) V (z A x) V (z A y)

Now we can use the algorithm from part (a) to determine, in polynomial
time, whetherthe resulting DNF formula is satisfiable. We havejust solved
3Sat in polynomial time. Since 3Sat is NP-hard, we must conclude that
P=NP!

4. The problem AllOrNothing3Sat asks, given a 3CNF boolean formula,
whether there is an assignment to the variables such that each clause either
has three True literals or has three False literals.

(a) Describe a polynomial-time algorithm to solve AllOrNothing3Sat.

(b) But 3Sat is NP-hard! Why doesn’t the existence of this algorithm prove
that P=NP?

5. (a) Suppose you are given a magic black box that can determine in poly¬

nomial time, given an arbitrary weighted graph G, the length of the
shortest Hamiltonian cycle in G. Describe and analyze a polynomial-
time algorithm that computes, given an arbitrary weighted graph G,
the shortest Hamiltonian cycle in G, using this magic black box as a
subroutine.

413

12. NP-Hardness

(b) Suppose you are given a magic black box that can determine in poly¬
nomial time, given an arbitrary graph G, the number of vertices in the
largest complete subgraph of G. Describe and analyze a polynomial¬
time algorithm that computes, given an arbitrary graph G, a complete
subgraph of G of maximum size, using this magic black box as a subrou¬
tine.

(c) Suppose you are given a magic black box that can determine in poly¬
nomial time, given an arbitrary graph G, whether G is 3 -colorable.
Describe and analyze a polynomial-time algorithm that either com¬
putes a proper 3 -coloring of a given graph or correctly reports that no
such coloring exists, using the magic black box as a subroutine. [Hint:
The input to the magic black box is a graph. Just a graph. Vertices and
edges. Nothing else.]

(d) Suppose you are given a magic black box that can determine in polyno¬
mial time, given an arbitrary boolean formula <f>, whether is satisfiable.
Describe and analyze a polynomial-time algorithm that either computes
a satisfying assignment for a given boolean formula or correctly reports
that no such assignment exists, using the magic black box as a subroutine.

(e) Suppose you are given a magic black box that can determine in polyno¬

mial time, given an arbitrary set X of positive integers, whether X can
be partitioned into two sets A and B such that Describe and

analyze a polynomial-time algorithm that either computes an equal
partition of a given set of positive integers or correctly reports that no
such partition exists, using the magic black box as a subroutine.

* v (f) Suppose you are given a magic black box that can determine in polyno¬
mial time, given an arbitrary extended regular expression R, whether R
matches any string. Describe and analyze a polynomial-time algorithm
that either finds a single string that matches a given extended regular
expression or correctly reports that no such string exists, using the magic
black box as a subroutine.

6. (a) Describe a polynomial-time reduction from Partition to SubsetSum.

(b) Describe a polynomial-time reduction from SubsetSum to Partition.

7. (a) Describe a polynomial-time reduction from UndirectedHamiltonian-

Cycle to DirectedHamiltonianCycle.

(b) Describe a polynomial-time reduction from DirectedHamiltonian¬
Cycle to UndirectedHamiltonianCycle.

8. (a) Describe a polynomial-time reduction from the HamiltonianPath prob¬

lem to HamiltonianCycle.

414

Exercises

(b) Describe a polynomial-time reduction from the HamiltonianCycle
problem to HamiltonianPath. [Hint: A polynomial-time reduction can
call the black-box subroutine more than once, but it doesn’t have to.]

9. Prove that PlanarCircuitSat is NP-hard. [Hint: Construct a gadget for
crossing wires.]

to. Prove that NotAllEqual3Sat is NP-hard.

11. Prove that 1-IN-3SAT is NP-hard.

12. Consider the following subtle variants of CNFSat. For each problem, the
input is a boolean formula <f> in conjunctive normal form, and the goal is to
determine whether <f> has a satisfying assignment.

(a) Suppose every clause of $ contains at most three literals and each
variable appears in at most three clauses. Prove that this variant of
CNFSat is NP-hard.

(b) Suppose every clause of $ contains exactly three literals and each
variable appears in at most four clauses. Prove that this variant of 3Sat
is NP-hard. [Hint: Solve part (a) first.]

r (c) Suppose every clause of $ can contain any number of literals, but each
variable appears in at most two clauses. Describe a polynomial-time
algorithm for this variant of CNFSat.

Y (d) Suppose every clause of $ contains exactly three literals and each variable
appears in at most three clauses. Prove that <f> must be satisfiable. (So
this variant of 3Sat is completely trivial!)

13. A boolean formula in exclusive-or conjunctive normal form (XCNF) is a
conjunction (And) of several clauses, each of which is the exclusive-or of
several literals; that is, a clause is true if and only if it contains an odd number
of true literals. The XCNF-Sat problem asks whether a given XCNF formula
is satisfiable. Either describe a polynomial-time algorithm for XCNF-Sat or
prove that it is NP-hard.

14. (a) Using the gadget in Figure 12.23(a), prove that deciding whether a given

planar graph is 3-colorable is NP-hard. [Hint: Show that the gadget
can be 3-colored, and then replace any crossings in a planar embedding
with the gadget appropriately.]

(b) Using part (a) and the gadget in Figure 12.23(b), prove that deciding
whether a planar graph with maximum degree 4 is 3-colorable is NP-hard.
[Hint: Replace any vertex with degree greater than 4 with a collection
of gadgets connected so that no degree is greater than four.]

415

12. NP-Hardness

Figure 12.23. (a) Gadget for planar 3-colorability. (b) Gadget for degree-4 planar 3-colorability.

15. Prove that the following problems are NP-hard.

(a) Given an undirected graph G, does G contain a simple path that visits
all but 17 vertices?

(b) Given an undirected graph G, does G have a spanning tree in which
every node has degree at most 23?

(c) Given an undirected graph G, does G have a spanning tree with at most
42 leaves?

(d) Given an undirected graph G = (V, £), what is the size of the largest
subset of vertices S c V such that at most 374 edges in E have both
endpoints in S?

(e) Given an undirected graph G = (V,E), what is the size of the largest
subset of vertices S ^ V such that each vertex in S has at most 473
neighbors in S?

(f) Given an undirected graph G, is it possible to color the vertices of G with
three different colors, so that at most 31337 edges have both endpoints
the same color?

16. Prove that the following variants of the minimum spanning tree problem

are NP-hard.

(a) Given a graph G, compute the maximum-diameter spanning tree of G.
(The diameter of a spanning tree T is the length of the longest path
in T.)

(b) Given a graph G with weighted edges, compute the minimum-weight
depth-first spanning tree of G.

(c) Given a graph G with weighted edges and a subset S of vertices of G,
compute the minimum-weight spanning tree all of whose leaves are in S.

(d) Given a graph G with weighted edges and an integer l, compute the
minimum-weight spanning tree with at most £ leaves.

Exercises

(e) Given a graph G with weighted edges and an integer A, compute the
minimum-weight spanning tree where every node has degree at most A.

17. There’s something special about the number 3.

(a) Describe and analyze a polynomial-time algorithm for 2Partition. Given
a set S of 2 n positive integers, your algorithm will determine in polyno¬
mial time whether the elements of S can be split into n disjoint pairs
whose sums are all equal.

(b) Describe and analyze a polynomial-time algorithm for 2Color. Given an
undirected graph G, your algorithm will determine in polynomial time
whether G has a proper coloring that uses only two colors.

(c) Describe and analyze a polynomial-time algorithm for 2Sat. Given a
boolean formula <f> in conjunctive normal form, with exactly two literals
per clause, your algorithm will determine in polynomial time whether <f>
has a satisfying assignment. [Hint: This problem is strongly connected
to topics described in an earlier chapter.]

18. There’s nothing special about the number 3.

(a) The problem 12Partition is defined as follows: Given a set S of 12 n
positive integers, determine whether the elements of S can be split into
n subsets of 12 elements each whose sums are all equal. Prove that
12Partition is NP-hard. [Hint: Reduce from 3Partition. It may be
easier to consider multisets first.]

(b) The problem 12C0LOR is defined as follows: Given an undirected graph G,
determine whether we can color each vertex with one of twelve colors,
so that every edge touches two different colors. Prove that 12C0LOR is
NP-hard. [Hint: Reduce from 3C0LOR.]

(c) The problem 12SAT is defined as follows: Given a boolean formula
<f> in conjunctive normal form, with exactly twelve literals per clause,
determine whether $ has a satisfying assignment. Prove that 12 Sat is
NP-hard. [Hint: Reduce from 3 Sat.]

19. (a) Describe a polynomial-time reduction from 3Sat to 4Sat.

(b) Describe a polynomial-time reduction from 4Sat to 3Sat.

*20. Describe a direct polynomial-time reduction from 4C0LOR to 3C0LOR. (This

is a lot harder than the opposite direction.)

417

12. NP-Hardness

21. A domino is a 1 x 2 rectangle divided into two squares, each of which is
labeled with an integer. 25 In a legal arrangement of dominoes, the dominoes
are lined up end-to-end so that the numbers on adjacent ends match.

For each of the following problems, either describe a polynomial-time
algorithm or prove that the problem is NP-hard:

(a) Given an arbitrary bag D of dominoes, is there a legal arrangement of
all the dominos in D?

(b) Given an arbitrary bag D of dominoes, is there a legal arrangement of a
dominos from D in which every integer between 1 and n appears exactly
twice?

Figure 12.24. A legal arrangement of dominos in which every integer between 0 and 6 appears twice

v (c) Given an arbitrary bag D of dominoes, what is the largest number of
dominos we can take from D to make a legal arrangement?

22. Pebbling is a solitaire game played on an undirected graph G, where each
vertex has zero or more pebbles. A single pebbling move consists of removing
two pebbles from a vertex v and adding one pebble to an arbitrary neighbor
of v. (Obviously, the vertex v must have at least two pebbles before the move.)
The PebbleDestruction problem asks, given a graph G = (V, E) and a
pebble count p(v) for each vertex v, whether is there a sequence of pebbling
moves that removes all but one pebble. Prove that PebbleDestruction is
NP-hard.

23. Recall that a 5-coloring of a graph G is a function that assigns each vertex
of G a “color” from the set { 0 , 1 , 2 , 3 , 4 }, such that for any edge uv, vertices u
and v are assigned different “colors”. A 5-coloring is careful if the colors
assigned to adjacent vertices are not only distinct, but differ by more than 1
(mod 5 ). Prove that deciding whether a given graph has a careful 5 -coloring
is NP-hard. [Hint: Reduce from the standard 5C0LOR problem.]

24. (a) A subset S of vertices in an undirected graph G is half-independent if

each vertex in S is adjacent to at most one other vertex in S. Prove that
finding the size of the largest half-independent set of vertices in a given
undirected graph is NP-hard.

25 These integers are usually represented by pips, exactly like dice. On a standard domino, the
number of pips on each side is between o and 6, although one can buy sets with up to 9 or even
12 pips on each side; we will allow arbitrary integer labels. A standard set of dominoes contains
exactly one domino for each possible unordered pair of labels; we do not assume that the inputs
to our problems have this property.

Exercises

Figure 12.25. A careful 5-coloring.

(b) A subset S of vertices in an undirected graph G is sort-of-independent if
if each vertex in S is adjacent to at most 374 other vertices in S. Prove
that finding the size of the largest sort-of-independent set of vertices in
a given undirected graph is NP-hard.

(c) A subset S of vertices in an undirected graph G is almost independent if at
most 374 edges in G have both endpoints in S. Prove that finding the size
of the largest almost-independent set of vertices in a given undirected
graph is NP-hard.

25. A subset S of vertices in an undirected graph G is triangle-free if, for every
triple of vertices u,v,w e S, at least one of the three edges uv, uw, vw is
absent from G. Prove that finding the size of the largest triangle-free subset
of vertices in a given undirected graph is NP-hard.

Figure 12.26. A triangle-free subset of 7 vertices. This is not the largest triangle-free subset in this
graph.

26. Let G = (V, E) be a graph. A dominating set in G is a subset S of the vertices
such that every vertex in G is either in S or adjacent to a vertex in S. The
DominatingSet problem asks, given a graph G and an integer k as input,
whether G contains a dominating set of size k. Prove that this problem is
NP-hard.

27. The RectangleTiling problem is defined as follows: Given one large
rectangle and several smaller rectangles, determine whether the smaller
rectangles can be placed inside the large rectangle with no gaps or overlaps.

(a) Prove that RectangleTiling is NP-hard.

(b) Prove that RectangleTiling is strongly NP-hard.

419

12. NP-Hardness

Figure 12.27. A dominating set of size 3 in the Peterson graph.

I 1

Figure 12.28. A positive instance of the RectangleTiling problem.

28. Let G be an undirected graph with weighted edges. A heavy Hamiltonian
cycle is a cycle C that passes through each vertex of G exactly once, such that
the total weight of the edges in C is more than half of the total weight of all
edges in G. Prove that deciding whether a graph has a heavy Hamiltonian
cycle is NP-hard.

Figure 12.29. A heavy Hamiltonian cycle. The cycle has total weight 34; the graph has total weight 67.

29. (a) A tonian path in a graph G is a path that goes through at least half of

the vertices of G. Show that determining whether a graph has a tonian
path is NP-hard.

(b) A tonian cycle in a graph G is a cycle that goes through at least half of
the vertices of G. Show that determining whether a graph has a tonian
cycle is NP-hard. [Hint: Use part (a). Or not.]

30. (a) A subset B of vertices in a graph G is a Burr set if removing every vertex

in B from G leaves a subgraph that does not contain a Hamiltonian path.
Prove that finding the smallest Burr set in a given graph is NP-hard.

420

Exercises

(b) A subset S of vertices in a graph G is a Schuyler set if removing every
vertex in S from G leaves a subgraph that does contain a Hamiltonian
path. Prove that finding the smallest Schuyler set in a given graph is
NP-hard.

31. This exercise asks you to prove that a certain reduction from VertexCover
to SteinerTree is correct. Suppose we want to find the smallest vertex
cover in a given undirected graph G = (V,E). We construct a new graph
H = ( V', E') as follows:

. V' = VUEU{z}

• E' = {ve | v e V is an endpoint of e € W} U {vz | v e V }.

Equivalently, we construct H by subdividing each edge in G with a new
vertex, and then connecting all the original vertices of G to a new apex
vertex z.

Prove that G has a vertex cover of size k if and only if there is a subtree
of H with k + |£| + 1 vertices that contains every vertex in E U {z}.

32. There are two different versions of the Hamiltonian cycle problem, one for
directed graphs and one for undirected graphs. Earlier in this chapter you
can find two proofs that the directed Hamiltonian cycle problem is NP-hard.

(a) Describe a polynomial-time reduction from the undirected Hamiltonian
cycle problem to the directed Hamiltonian cycle problem. Prove your
reduction is correct.

(b) Describe a polynomial-time reduction from the directed Hamiltonian
cycle problem to the undirected Hamiltonian cycle problem. Prove your
reduction is correct.

(c) Which of these two reductions implies that the undirected Hamiltonian
cycle problem is NP-hard?

33. For each of the following problems, either describe a polynomial-time
algorithm or prove that the problem is NP-hard.

(a) A double-Eulerian tour in an undirected graph G is a closed walk that
traverses every edge in G exactly twice. Given a graph G, does G have a
double-Eulerian tour?

(b) A double-Hamiltonian tour in an undirected graph G is a closed walk
that visits every vertex in G exactly twice. Given a graph G, does G have
a double-Hamiltonian tour?

(c) A double-Hamiltonian circuit in an undirected graph G is a closed walk
that visits every vertex in G exactly twice and traverses each edge in G at
most once. Given a graph G, does G have a double-Hamiltonian circuit?

421

12. NP-Hardness

(d) A triple-Eulerian tour in an undirected graph G is a closed walk that
traverses every edge in G exactly three times. Given a graph G, does G
have a triple-Eulerian tour?

(e) A triple-Hamiltonian tour in an undirected graph G is a closed walk that
visits every vertex in G exactly three times. Given a graph G, does G
have a triple-Hamiltonian tour?

34. Consider the following solitaire game. The puzzle consists of an n x m
grid of squares, where each square may be empty, occupied by a red stone,
or occupied by a blue stone. The goal of the puzzle is to remove some of
the given stones so that the remaining stones satisfy two conditions: (1)
every row contains at least one stone, and (2) no column contains stones of
both colors. For some initial configurations of stones, reaching this goal is
impossible.

0

O

O

O

£3

S3

S3

O

O

O

S3

S3

O

O

O

O

S3

S3

S3

O

O

O

K

S3

S3

O

O

S3

O

O

S3

S3

A solvable puzzle and one of its many solutions. An unsolvable puzzle.

Prove that it is NP-hard to determine, given an initial configuration of
red and blue stones, whether the puzzle can be solved.

35. Each of the following games involves an n x m grid of squares, where each
square is either empty or occupied by a stone. In a single move, you can
remove all the stones in an arbitrary column.

(a) Prove that it is NP-hard to find the smallest subset of columns that can
be cleared so that at most one stone remains in each row of the grid.

(b) Prove that it is NP-hard to find the largest subset of columns that can be
cleared so that at least one stone remains in each row of the grid.

*(c) Prove that it is NP-hard to determine whether any subset of columns can
be cleared so that exactly one stone remains in each row of the grid.

36. Jeff tries to make his students happy. At the beginning of class, he passes out
a questionnaire that lists a number of possible course policies in areas where
he is flexible. Every student is asked to respond to each possible course
policy with one of “strongly favor”, “mostly neutral”, or “strongly oppose”.
Each student may respond with “strongly favor” or “strongly oppose” to at
most five questions. Because Jeff’s students are very understanding, each

422

Exercises

student is happy if (but only if) he or she prevails in just one of his or her
strong policy preferences. Either describe a polynomial-time algorithm for
setting course policy to maximize the number of happy students, or show
that the problem is NP-hard.

37. You’re in charge of choreographing a musical for your local community
theater, and it’s time to figure out the final pose of the big show-stopping
number at the end. (“Streetcar!”) You’ve decided that each of the n cast
members in the show will be positioned in a big line when the song finishes,
all with their arms extended and showing off their best spirit fingers.

The director has declared that during the final flourish, each cast member
must either point both their arms up or point both their arms down; it’s your
job to figure out who points up and who points down. Moreover, in a fit of
unchecked power, the director has also given you a list of arrangements that
will upset his delicate artistic temperament. Each forbidden arrangement
is a subset of the cast members paired with arm positions; for example:
“Marge may not point her arms up while Ned, Apu, and Smithers point their
arms down.”

Prove that finding an acceptable arrangement of arm positions is NP-hard.

38. The next time you are at a party, one of the guests will suggest everyone play a
round of Three-Way Mumbletypeg, a game of skill and dexterity that requires
three teams and a knife. The official Rules of Three-Way Mumbletypeg
(fixed during the Holy Roman Three-Way Mumbletypeg Council in 1625)
require that (1) each team must have at least one person, (2) any two people
on the same team must know each other, and (3) everyone watching the
game must be on one of the three teams. Of course, it will be a really fun
party; nobody will want to leave. There will be several pairs of people at
the party who don’t know each other. The host of the party, having heard
thrilling tales of your prowess in all things algorithmic, will hand you a list
of which pairs of party-goers know each other and ask you to choose the
teams, while he sharpens the knife.

Either describe and analyze a polynomial time algorithm to determine
whether the party-goers can be split into three legal Three-Way Mumbletypeg
teams, or prove that the problem is NP-hard.

39. The party you are attending is going great, but now it’s time to line up for
The Algorithm March (T X" ') X X X y L AJ! This dance was originally
developed by the Japanese comedy duo Itsumo Kokokara (V ^ O Z X h* <o)
for the children’s television show PythagoraSwitch ( X 9 r Ty X J "j X).
The Algorithm March is performed by a line of people; each person in line
starts a specific sequence of movements one measure later than the person

423

12. NP-Hardness

directly in front of them. Thus, the march is the dance equivalent of a
musical round or canon, like “Row Row Row Your Boat”.

Proper etiquette dictates that each marcher must know the person
directly in front of them in line, lest a minor mistake during lead to horrible
embarrassment between strangers. Suppose you are given a complete list
of which people at your party know each other. Prove that it is NP-hard
to determine the largest number of party-goers that can participate in the
Algorithm March. You may assume without loss of generality that there are
no ninjas at your party.

*40. Prove that the following problems about nondeterministic finite-state au¬
tomata and regular expressions are NP-hard:

(a) Given an NFA M over the alphabet E = { 0 , 1 }, is there a string in E* that
M does not accept?

(b) Given an acyclic NFA M over the alphabet E = { 0 , 1 }, what is the length
of the shortest string in E* that M does not accept?

(c) Given a regular expression R over the alphabet E = { 0 , 1 }, is there a
string in E* that R does not match?

(d) Given a star-free regular expression R over the alphabet E = { 0 , 1 }, what
is the length of the shortest string in E* that R does not match?

(In fact, problems (a) and (c) are PSPACE-complete; even proving that these
problems are in PSPACE is nontrivial.)

*41. (a) Describe a polynomial-time algorithm for the following problem: Given
an NFA M over the alphabet E = { 0 , 1 }, is there a string in E* that M
does accept?

(b) Describe a polynomial-time algorithm for the following problem: Given
a regular expression R over the alphabet E = { 0 , 1 }, is there a string in
E* that R does match?

(c) The complement of any regular language is another regular language.
So why don’t these two algorithms, together with the NP-hardness results
in Problem 40, prove that P=NP?

42. Charon needs to ferry n recently deceased people across the river Acheron
into Hades. Certain pairs of these people are sworn enemies, who cannot
be together on either side of the river unless Charon is also present. (If two
enemies are left alone, one will steal the obol from the other’s mouth, leaving
them to wander the banks of the Acheron as a ghost for all eternity. Let’s

424

Exercises

just say this is a Very Bad Thing.) The ferry can hold at most k passengers
at a time, including Charon, and only Charon can pilot the ferry. 26

Prove that it is NP-hard to decide whether Charon can ferry all n people
across the Acheron unharmed (aside from being, you know, dead). The
input for Charon’s problem consists of the integers k and n and an n-vertex
graph G describing the pairs of enemies. The output is either True or False.

Please do not write your solution in classical Latin.

26 This is a generalization of the well-known wolf-goat-and-cabbage puzzle, whose first known
appearance is in the remarkable medieval manuscript Propositiones ad Acuendos Juvenes [Problems
to Sharpen the Young ].

XVIII. Propositio De Homine et Capra et Lvpo.

Homo quidam debebat ultra fluuium transferre lupum, capram, et fasciculum cauli. Et
non potuit aliam nauem inuenire, nisi quae duos tantum ex ipsis ferre ualebat. Praeceptum
itaque ei fuerat, ut omnia haec ultra illaesa omnino transferret. Dicat, qui potest, quomodo
eis illaesis transire potuit?

Solutio. Simili namque tenore ducerem prius capram et dimitterem foris lupum et caulum.

Turn deinde uenirem, lupumque transferrem: lupoque foris misso capram naui receptam ultra
reducerem; capramque foris missam caulum transueherem ultra; atque iterum remigassem,
capramque assumptam ultra duxissem. Sicque faciendo facta erit remigatio salubris, absque
uoragine lacerationis.

For those few readers whose classical Latin is a little rusty, here is an English translation:

XVIII. The Problem of the Man, the Goat, and the Wolf.

A man needed to transfer a wolf, a goat, and a bundle of cabbage across a river. However,
he found that his boat could only bear the weight of two [objects at a time, including the
manj. And he had to get everything across unharmed. Tell me if you can: How they were
able to cross unharmed?

Solution. In a similar fashion [as an earlier problem], I would first take the goat across
and leave the wolf and cabbage on the opposite bank. Then I would take the wolf across;
leaving the wolf on shore, I would retrieve the goat and bring it back again. Then I would
leave the goat and take the cabbage across. And then I would row across again and get the
goat. In this way the crossing would go well, without any threat of slaughter.

The most likely author of the Propositiones is the prolific 8th-century English scholar Alcuin of
York. The evidence for Alcuin’s authorship of this treatise is somewhat circumstantial; however,
we do know from his correspondence with Charlemagne that he sent the emperor some “simple
arithmetical problems for fun”. Most modern scholars believe that even if Alcuin did write the
Propositiones, he did not invent all of the problems himself, but rather collected them from even
earlier sources.

Some things never change.

425

A wisely chosen illustration is almost essential to fasten the truth upon the
ordinary mind, and no teacher can afford to neglect this part of his preparation.

— Howard Crosby (c.1880)

One showing is worth a hundred sayings.

- Alan Watts (misquoting a Chinese proverb), The Way of Zen (1957)

Please do not think that this is a neutral matter and that the only advantage of
doing without pictures is that of saving space. Pictures in textbooks actually
interfere with the learning process.

- Neville Martin Gwynne, Gwynne's Grammar (2013)

Image Credits

All figures in this book are original works of the author, except for the following.
All listed works are in the public domain unless otherwise indicated.

• Figure 0.1 (page 5) — Biblioteca nazionale Braidense (Milano)
http://atena.beic.it/webclient/DeliveryManager 7 pid = 2953344

• Figure 0.2 (page 5) — Internet Archive
https://archive.org/details/archimedisoperao5eutogoog/page/n377

• Figure 1.14 (page 45) — Internet Archive
https://archive.org/details/p1rcrationsmoolucauoft/page/162

• Figure 1.23 (page 60) — Derived from a crayon portrait of the author by
Tina Erickson (2000); included with permission of the artist.

• Figure 5.1 (page 186) — Wikimedia Commons

https //commons .wikimedia .org/wiki/File: Tabula_Peutingeriana_-_Miller.jpg

Figure 5.2 (page 187) — Gallery of “Legal Trees” published by the Yale Law
Library under a Creative Commons Licence

https://www.flickr.com/photos/yalelawlibrary/albums/72157621954683764

Figure 5.3 (page 187) — Internet Archive
https://archive.0rg/details/A077240124/page/n261

Exercises 5.20 (page 214) and 8.21 (page 302) — Original puzzles by the
author, inspired by Jason Batterson and Shannon Rogers, Beast Academy
Math: Practice 3A, 2012.

https://beastacademy.c0m/pdf/3A/printables/AngleMazes.pdf

https://www.beastacademy.com/resources/printables.php

Figure 10.1 (page 326) — T[homas] E. Harris and F[rank] S. Ross. Fundamen¬
tals of a method for evaluating rail net capacities. The RAND Corporation,
Research Memorandum RM-1517, October 24, 1955. U. S. Government work
in the public domain.

http://www.dtic.mil/dtic/tr/fulltext/u2/093458.pdf

7. Have something to say.

2. Say it.

j. Stop when you have said it.

4. Give the paper a proper title.

— John Shaw Billings, "An Address on Our Medical Literature",
International Medical Congress, London (1881)

You know, I could write a book.

And this book would be thick enough to stun an ox.

— Laurie Anderson, "Let X=X", Big Science (1982)

Colophon

This book was edited in TeXShop and typeset using pdfKTpX (MacTeX-2018) using
the memoir document class (with madsen chapter style, komalike head style,
and Ruled page style); several standard packages including amsmath, babel,
enumitem, mathdesign, microtype, and standalone; and an embarrassing
amount of customization and TgX-ly\Xing. The text is typeset in Bitstream
Charter, Apte/rioia, Roboto, and Inconsolata. Except as indicated in the
Image Credits, all figures were drawn by the author using OmniGraffle Pro
and included as PDF files, using the graphicx ETgX package. Portions of our
programming have been mechanically reproduced, and we now conclude our
broadcast day.